N00b to C++ here, have some general questions.

So I've been taking a C++ in school and sometimes it seems like I've to turn to sources other than my prof for help. Hence after some googling, I'm here. Now I have a few questions.

I've been reading my text and prof's lecture notes, and there are somethings that I can't get my mind around.

1)

Code:

---

void printClassInfo( void )

---

What exactly does the void in brackets do? Would you need the whole program to answer the question? I know what something like " void displayMessage( string courseName ) " does [sorry for not posting it in code, I figured it would "break the layout"].

2) How exactly does the getline command work? An example to understand it's working would be very helpful, thanks in advance.

3) What is the purpose of something like "using namespace std;" or "using std: :cout;" or "using std: :cin" etc?

Any help on these three questions would be really helpful, and hopefully play an important role when the class advances further. Also, I have no previous exposure to C - I know it's a handicap, but I hope that it's not something that'll cripple me. Thanks in advance, guys/gals!

1. The void inside the ()'s shows what the function accepts. It accepts void, which means nothing. What is the difference between leaving out the void and not? Ah, this is going to be good confusion for you. :p
   
   For C, if you use empty ()'s in a function prototype, it means the actual parameter list to that function is undefined. It means you can call it with anything you want and the function just goes. The function itself can be defined to accept anything. This can lead to issues at runtime, so this is generally not how variable length args are done anymore.
   
   In C++ and the latest dialect of C, which is C99, as far as I know, there is no technical difference between (void) and (), but I might be totally wrong on that. C99 might still make the difference... actually I think it's militant on the difference. :D
   
   For when it doesn't matter (in C++), it's good to include the void since it makes your intentions explicit, but it's not necessary, and can be left out. Just make sure you remember exactly what language and dialect of said language you're using, and what you really intend.
   
2. With regards to what? You said you're quite new, so I don't think you want an extremely detailed level of explanation, but I shall attempt to throw a bunch of information at you anyway and you can take whatever you are able to understand. :)
   
   On most systems, and in C (and subsequently C++), there are 3 files opened by default:
   
   
   • stdin
   • stdout
   • stderr
   
   
   All three files are usually refer in some ways to the console. Console input and output are handled by means of these "files". When you print information with cout, it's actually writing that data to the file stdout, which stands for Standard Out. Similary, stdin refers to Standard In. getline() simply reads a bunch of chars from this "file", possibly one by one.
   
   This can get quite complicated, so for now, just accept it works, and get very used to it.
   
3. To avoid naming conflicts, the concept of namespaces was incorporated into C++. This means you can take classes, functions, and variables and put them in a namespace. There's a namespace named std. That namespace includes a bunch of basic things you use all the time, like cout and cin. To use them you would do something like this:
   
   
   Code:

   ---

   std::cout << "hello world" << std::endl;

   ---

   Since typing std:: can be tiresome, you can tell the compiler to automatically check out certain namespaces or certain elements inside certain namespaces when it comes across an unknown element.
   
   
   Code:

   ---

   using namespace std;

...

cout << "hello world" << endl;

   ---

   When cout and endl are reached by the compiler, it knows to check namespace std to see if cout and endl happen to be there. If it finds them there, it uses those particular elements.
   
   Again, namespaces are there so you can avoid naming conflicts. You might not make a class with the name cout, given that you know one exists, but you might accidentally make one for something less well known and that might cause issues. Namespaces allows you to organize your code.

Quote:

---

Originally Posted by Bash

So I've been taking a C++ in school and sometimes it seems like I've to turn to sources other than my prof for help. Hence after some googling, I'm here. Now I have a few questions.

I've been reading my text and prof's lecture notes, and there are somethings that I can't get my mind around.

1)

Code:

---

void printClassInfo( void )

---

What exactly does the void in brackets do? Would you need the whole program to answer the question? I know what something like " void displayMessage( string courseName ) " does [sorry for not posting it in code, I figured it would "break the layout"].

---

In C++, the void in brackets is not NECESSARY, but it is a way to indicate explicitly that "this function takes no parameters". In C there is a difererence between a function declared as

Code:

---

void printClassInfo()

---

and

Code:

---

void printClassInfo(void)

---

- the first one has "unspecified parameters" (could be zero, one or ten - no way for the compiler to tell what is right and wrong of these). The second one has no parameters - that's it, the compiler will complain if you give it any parameter(s).

Quote:

---

2) How exactly does the getline command work? An example to understand it's working would be very helpful, thanks in advance.

---

Technically, it's not a "command", but a function.

Essentially, it reads data into a C-string (char array) from the input stream (usually cin) until it reaches a newline (or some other delimiter specified by the third parameter) - or it reaches the size specified.

There is also a getline that operates on a C++ std::string type - this is not part of the iostream member functions, but it essentially does the same thing - you pass a stream as one of the parameters instead.

Quote:

---

3) What is the purpose of something like "using namespace std;" or "using std: :cout;" or "using std: :cin" etc?

---

When working on very large projects, you sometimes get what's called "name clashes", where two programmers decide to call a function the same thing. By using different namespaces, these name-clashes can be limited to a smaller portion of code. A namespace may be a functional unit. E.g. the graphics part of an OS may have it's own namespace, the kernel has another namespace.

In a large application that does order, stock and invoicing, the ordering system has one namespace, the stock counting another, and the invoicing part a third namespace. That means that the variabl "current_customer" in the ordering system is a distinct different variable than "current_customer" in the invoice system, for example - because one is, when you write it's full name is "order::current_customer" and the other "invoice::current_customer".

You may find that there is a "sum_customer_order" in both the ordering system and in the invoice system - but they do different things: in the ordering system, it calculates the total of the order as the customer orders. In the invoice stage, you only invoice for the items actually delivered, and any back-order items will not be included in this stage - they are invoiced when they are delivered.

Quote:

---

Any help on these three questions would be really helpful, and hopefully play an important role when the class advances further. Also, I have no previous exposure to C - I know it's a handicap, but I hope that it's not something that'll cripple me. Thanks in advance, guys/gals!

---

Thanks for both the replies, MacGyver and matsp :) I now have an understanding of it and pray they don't go away.

About the getline function - can you use it to store a string in an array? For example, let's say I input 04953045, can I use the getline function to store it in an array of length l-1 (here's it's 8, duh! :P) so that I can actually have different numbers in a[0],a[1],etc? If so, how would I do it?

[Oh, btw, everytime I hear that something's a function, I end up thinking f(x) in math and my mind can't grasp the meaning quite well ={]

Quote:

---

Originally Posted by Bash

About the getline function - can you use it to store a string in an array? For example, let's say I input 04953045, can I use the getline function to store it in an array of length l-1 (here's it's 8, duh! :P) so that I can actually have different numbers in a[0],a[1],etc? If so, how would I do it?

---

istream::getline stores the input as characters in a char array, so cin.getline(a, 10) would give you an array of a[0] = '0', a[1] = '4', ... a[7] = '5', a[8] = '\0'.

Quote:

---

[Oh, btw, everytime I hear that something's a function, I end up thinking f(x) in math and my mind can't grasp the meaning quite well ={]

---

Yes, mathematical and programming functions are generally not the same thing. In some languages, such as Pascal, there are "procedures" and "functions". A procedure is the same as a C function that returns nothing (void function), and a "function" in Pascal is the same as something that returns a value (but not necessarily a numerical value - it could be "anything").

But it's best if you "program your brain" to think of functions in programming and that they aren't the same as functions in mathematics.

--
Mats

I don't see why the math model of functions should be confusing when applied to programming.

Take something simple: f(x) = x + 5

It accepts a parameter x, and returns (x + 5). You can program something similar quite easily:

Code:

---

int f(int x)
{
        return x + 5;
}

---

Oh, sure math functions are WITHIN the functions in programming - it's just that math functions are much more restricted than programming functions.

--
Mats

Quote:

---

So I've been taking a C++ in school and sometimes it seems like I've to turn to sources other than my prof for help.

---

This is normal, especially if it's your first programming class. The concepts are new, the terminology is new, and (assuming this is a university class) you will be covering lots of ground very quickly.

And, it will probably require more "homework time" than your typical class. It's not that you will be assigned lots of homework, but you will frequently "get stuck" on some small issue or bug, and that can eat-up many hours. It would be a good idea to get into a study group (if that's not considered "cheating").

So I have this homework assignment that's due on Friday, and I had started working on it. Basically what we have to do is accept a number (like 95468) and write a program that outputs the sum of the digits, the biggest digit, the smallest digit, the first digit and the last digit. So far I've figured that if I use modulus and keep dividing by ten, I can separate teh values. But for that, the only way I can write it is to have a set number of digits (like it has to be a 5-digit number). Overall it doesn't seem to work right - I shall post the code and any way to optimize it or any suggestions would be helpful.

There is also a second part to the question - It's the "Press Y to continue or N to quit" The program would also quit if you entered -99999. To do this, would I envelop the whole body of it in an if-else loop? Where would I place it?

Code:

Code:

---

#include <iostream>
#include <string>
using namespace std;

int main ()
{
        int iX;
        int iSmall;
        int iBig;
        int iY;
        int iSmallestDigit = 9;
        int iTestNum;
        int iDigit;
        int iLargestDigit = 0;
        int iSum = 0;
        int iSumDump;

        cout << "Class Information --"
                << "\n CIS 25 - ++ Programming"
                << "\n Laney College" << endl;

        cout << "\nAssigment Information --"
                << "\n Assigment Number: Lab 03 Exercise #1"
                << "\n Written by: Amol Mundayoor"
                << "\n Due Date: 09/26/07" << endl;

        cout << "\n\nEnter a 5-digit Integer: ";
        cin  >> iX;


                if (iX % 2 == 0)
                        {
                        cout << iX << " Is an even ";
                        }
                else
                        {
                        cout << iX << " Is an odd ";
                        }
                if (iX > 0)
                        {
                        cout << "and positive number\n";
                        }
                else if (iX < 0)
                        {
                        cout << "and negative number\n";
                        }
                else
                        {
                        cout << "Invalid Input" << endl;
                        }

                iSmall = iX % 10;
                        cout << "The least significant digit is " << iSmall <<endl;

                iBig = iX/10000;
                        cout << "The most significant digit is " << iBig << endl;

                iTestNum = iX;
                       
                do
                {
               
                iDigit = iTestNum % 10;
               
                if ( iSmallestDigit > iDigit )
                       
                        {
                                iSmallestDigit = iDigit;
                        }
               
                iTestNum /= 10;
               
                } while ( iTestNum );
       
                        cout << "Smallest Digit: " << iSmallestDigit << endl;
               
                iY = iX;

                do
                {
               
                iDigit = iY % 10;
               
                if ( iLargestDigit < iDigit )
                       
                        {
                                iLargestDigit = iDigit;
                        }
               
                iY /= 10;
               
                } while ( iY );
       
                        cout << "Largest Digit: " << iLargestDigit << endl;

               
                iSumDump = iX;       
                for (iSumDump = iX % 10; iX /= 10; iSum=iSum+iSumDump);
                        cout << "The sum of Digits is: "<< iSum << endl;


  return 0;
}

---

And my for loop doesn't seem to work either - where is it not logically right? =/ Thanks a bunch, at least I'm getting started here ^^;

You are right, the for-loop doesn't work right. You are not updating the "current digit" in iSumDump, and you are not counting the final (most significant) digit.

--
Mats

I was hoping to use iSumDump as a temp value. I wanted to keep storing the remainders of teh division in iSumDump and keep adding to iSum. For example,

If the number is 94602, I want iSumDump to have the value "2" in the first run, "0" in teh second, "6" in the third, etc. And at the end of the loop, before it gets out, I'd keep adding it to iSum (iSum would be updated with "2" and then the rest would be added on. Would such an operation work with a for loop or should I do something else, like a do...while?

Yes, and in your current implementation, you are taking the first digit into iSumDump, then adding that until your number divided by 10 is zero. So if you enter 12345, you add the number five four times to iSum.

There's nothing wrong with the variable as such, and if the above was what you WANT to do, then it would be a fine solution. However, you probably want to add 1 + 2 + 3 + 4 + 5 (or, rather, to make life easy, 5 + 4 + 3 + 2+ 1), not 5 + 5 + 5 + 5.

So you need to do two things:
1. Make sure that iSumUpdate is representing the current digit in the number, not always the least significant.
2. Make sure ALL digits are accumulated.

--
Mats

OHHHH, I see what I was doing wrong. After fiddling with the for loop, I couldn't come up with something that actually worked. It's kind of a bummer to me that my for loop didn't work, but I made a do while (it felt a LOT more simple to make one)

Code:

---

do
{
iSumDump = iX % 10;
iSum = iSumDump+iSum;
iX /= 10;
} while (iX);

---

I don't want to ask anyone to do anything for me, but I have no idea on how the right for loop would look like. Can anyone show the actual working loop? I know how to do it in do...while, but I'm totally lost when I do it in the for loop. Thanks a lot for all your help :)

The for loop would depend upon what you set iX to. Just that block you posted would look like something like this:

Code:

---

for(;iX;iX /= 10)
{
        iSumDump = iX % 10;
        iSum = iSumDump+iSum;
}

---

The main difference with for and while vs do-while is that a for/while loop will have its condition checked prior to entering the loop. On the other hand, a do-while will execute once regardless if the condition is true or not. This inherently means a do-while loop will always execute at least once.

That's interesting, because I've never come across a for-while loop before.

What if I wanted my program to accept an n-digit integer instead of a 5-digit integer (I didn't know how to solve it unless I assumed that the entered integer is a 5-digit one)? Would I still use
"istream::getline"? Or is there a way to convert the number to a string and then work with it?

No, I meant for loops and while loops vs do-while loops. Not that there was such a thing as a for-while loop.... lol.....

BTW, what about your program currently can't handle the n-digit scenario? You have a loop. It should work if your algorithm is correct (which I have some doubts about from looking at the loop).

Ok, so using for in the way MacGyver does is quite contrived (not that it's not a valid use, just that it's a bit "unusual" - a bit like using a screwdriver to open the tape on a box - not completely "strange", but not quite as neat as using a knife.

If you really want to, you could solve it with (but I recommend AGAINST this):

Code:

---

for(;iX; iSumDump = iX % 10, iSum += iSumDump, iX /= 10);

---

But I prefer either the do-while or while construction - the while construction also solves the case of a input value of zero not having to be calculated at all [not that it really matters in this case - but I always try to think about how to avoid executing code that isn't necessary to execute]

Code:

---

while(iX) {
  iSum += iX % 10;
  iX /= 10;
}

---

I removed the need for iSumDump, as it's not really necessary in this case. This is probably the simplest version (the same version can be done in a do-while loop of course, only differences is that iX is tested after the first iteration of the loop, rather than before).

As for the n-digit scenario, if we assume that n is not larger than about 10 or so, you can still do it with integer numbers, same way you are doing right now.

The only part I can see where you don't deal correctly with "any" number of digits is where you divide the number by 10000. Now, if you think about it, what method do you think you can use to find out what the most significant digit - perhaps some of your existing code already does that, of sorts...

--
Mats

I still haven't figured out the n-digit part, but I'll take care of it later. I was attempting to modify the program even more, for example, giving feedback when you enter a float I want my program to spew out "Invalid Input" or something similar. So basically, how would I check for floats? (The n-digit scenario here). I'd also appreciate it if the same could be done with character inputs (like abc34 or abed)

The best way to deal with validating input is to read a line of text and then verify that it can be converted correctly with no "leftovers".

Use for example

Code:

---

std::string str;
std::getline(cin, str);

---

You can then validate that you can convert the string to a number, e.g.

Code:

---

#include <sstream>

...
  stringstream ss;

  ss << str;    // Fill the stringstream with the string entered.

  if (ss >> someint) {
    ... valid input ...
  } else {
      ... input was not valid ...
  }

---

This doesn't attempt to recognise "not all of the input was consumed". This is much trickier to do, and I think it's not necessarily meaningfull to try to do that.

--
Mats

matsp, I'm sorry to say that what you said went way off the top of my head :p

After a lot of googling, I came across this code chunk, but I've bolded and underlined parts that I don't get:

Code:

---

#include <iostream>
#include <limits>

using namespace std;

int main() {
  int number = 0;
  cout << "Enter a integral value: ";
  cin >> number;
  cin.ignore(numeric_limits<int>::max(), '\n');
  if(number<0)
  {
    cout << "Error: You entered a negative number.";
    return 1;
  }
   
  if (!cin || cin.gcount() != 1)
    cout << "Not a numeric value.";
  else
    cout << "Your entered number: " << number;
    return 0;
}

---

Can someone help me understand what it does? Thanks a lot guys :)

Edit: Oh I just noticed the new header file, lol. Now I should be off figuring out what that one does, lol :D

That code isn't exactly complete, since it doesn't tell you how to clear the stream of errors and ask again. I prefer this method of checking for integers:

Code:

---

int number = 0;
while (!(cin >> number) || cin.get() != '\n')
{
    cin.clear();
    cin.ignore(numeric_limits<streamsize>::max(), '\n');
    // output your error message here and tell the user to try again.
}

---

Here's what it does

Code:

---

!(cin >> number)

---

If cin fails to read in an int, it goes into a fail state. For example, if the user types characters like "abc", that isn't an int. Checking the return value allows you to check for the fail state. So if cin >> number fails, then it will evaluate to false and !(cin >> number) will be true. So if it fails, we'll enter the loop to clear the error and try again.

Code:

---

|| cin.get() != '\n'

---

If the input doesn't fail, then we check the next character. If the user types a number and hits enter, then it should be a newline, which is '\n'. If it is not a newline, then the user typed extra characters. So for example if the user types "123.0" or "123abc", this second part will be true and the code will enter the error handling loop.

Code:

---

cin.clear();

---

This clears the failbit. If the loop was entered because the read failed, then this clears that flag so you can try to read again. If the fail bit is not set this does nothing, so it is safe to call here.

Code:

---

cin.ignore(numeric_limits<streamsize>::max(), '\n');

---

This ignores all characters (up to the newline) in the buffer. It gets rid of all the bad input so the user can try again with a clean bluffer. You could also use int instead of streamsize, or even cin.ignore(1000, '\n');. The difference is that 1000 will ignore a maximum of 1000 characters, so if the user types in more than 1000 characters, they won't all be ignored. Using the numeric_limits version makes sure you ignore them all just in case.

So overall, you try to read a number. If it fails or there are extra garbage characters, you clear and empty the cin stream and try again. You keep doing this in a loop until the user enters a number.

Code:

---

if (!cin || cin.gcount() != 1)

---

In that code, the !cin is like the !(cin >> number) I used, it just checks cin later. The cin.gcount != 1 is like the cin.get() != '\n', it checks to make sure there is only one character (the newline) left in the stream.

Oh man, I ran into so much trouble trying to come up with solutions as to why I couldn't just copy paste your code. Copy pasting is baaaaad, but in the end a lot of good came off of it as I knew more about what each part of each line does. Everything's uber fine now although it took me an hour to "perfectly adapt" :D

Glad it's working.

If you want you can post your final solution so others can benefit and in case there are any hidden gotchas we can point out.

Sorry for the late reply :). It'll be my pleasure to post it here.

Code:

---

#include <iostream>
#include <string>
#include <limits>
using namespace std;

int main ()
{
        int iX = 0;
        int iSmall;
        int iBig;
        int iY;
        int iSmallestDigit = 9;
        int iTestNum;
        int iDigit;
        int iLargestDigit = 0;
        int iSum = 0;
        int iSumDump;
        int iZ;
        char Check;

        cout << "Class Information --"
                << "\n CIS 25 - ++ Programming"
                << "\n Laney College" << endl;

        cout << "\nAssigment Information --"
                << "\n Assigment Number: Lab 03 Exercise #2"
                << "\n Written by: Amol Mundayoor"
                << "\n Due Date: 10/04/07" << endl;


        do
        {
       
        cout << "\n\nEnter a 5-digit Integer (or -999999 to exit): ";
       
        while (!(cin >> iX) || cin.get() != '\n')
        {
   
                cin.clear();
                cin.ignore(numeric_limits<int>::max(), '\n');
                cout << " ERROR! Invalid input, please try again: ";
       
        }       


        if (iX == -999999)
                        break;
                       
       
                if (iX &#37; 2 == 0)
                        {
                        cout << iX << " is an even ";
                        }
                else
                        {
                        cout << iX << " is an odd ";
                        }
                if (iX > 0)
                        {
                        cout << "and positive number\n";
                        }
                else if (iX < 0)
                        {
                        cout << "and negative number\n";
                        }
                else
                        {
                        cout << "Invalid Input" << endl;
                        }

                iSmall = iX % 10;
                        cout << "The least significant digit is " << iSmall <<endl;

                iBig = iX/10000;
                        cout << "The most significant digit is " << iBig << endl;

                iTestNum = iX;
                       
                do
                {
               
                iDigit = iTestNum % 10;
               
                if ( iSmallestDigit > iDigit )
                       
                        {
                                iSmallestDigit = iDigit;
                        }
               
                iTestNum /= 10;
               
                } while ( iTestNum );
       
                        cout << "Smallest Digit: " << iSmallestDigit << endl;
               
                iY = iX;

                do
                {
               
                iDigit = iY % 10;
               
                if ( iLargestDigit < iDigit )
                       
                        {
                                iLargestDigit = iDigit;
                        }
               
                iY /= 10;
               
                } while ( iY );
       
                        cout << "Largest Digit: " << iLargestDigit << endl;

               
                iZ = iX;
                iSum=0;
                do
                {
                        iSumDump = iZ % 10;
                        iSum = iSumDump+iSum;
                        iZ /= 10;
                } while (iZ);
                        cout << "The sum of Digits is: "<< iSum << endl;

               
                cout << "\n\nWould you like to work with another value? Press [n] for No and [y] for Yes:";
                cin >> Check;
               
        } while(Check == 'y' || Check == 'Y');


  return 0;
}

---

.

I've been reading, and I has a question. What's the difference between swap(); and void swap ();? If I write a code like this:

Code:

---

(a>b) ? a=b: swap();

---

Does it do this:

it checks if a is greater than b, and if it is, then it assigns the value of b to a, if not the values are swapped (what's in a goes to b, and what's in b goes to a). ? (I just figured out about the ternary conditional operator - I was testing if I knew how to use that too :p). Thanks guys :)

Edit:
Oh, and is the above code (the single line one) the same as this:

Code:

---

(a>b) ? a=b: swap(a,b);

---

Quote:

---

What's the difference between swap(); and void swap ();?

---

I have no context to work with, but "swap();" is probably a call to some function named swap that takes no arguments (or allows no arguments and uses default arguments instead). "void swap();" is the prototype of a function named swap that takes no arguments and does not return a value.

Quote:

---

it checks if a is greater than b, and if it is, then it assigns the value of b to a, if not the values are swapped (what's in a goes to b, and what's in b goes to a). ?

---

Probably not. It depends on what exactly this swap() function actually does, and whether a and b are member variables etc... but most likely you are trying to call a function that does not exist. The swap() function template from <algorithm> takes two arguments by reference and swaps them, assuming that they are of a type that is copy constructible and copy assignable.

So, your second version has the right idea, but this is not the place to use the ternary operator. Instead, write:

Code:

---

if (a > b)
{
    a = b;
}
else
{
    swap(a, b);
}

---

Well our prof had asked this question in class:

(1) Given that the following prototypes are given in some program. What would and may be potential problems from each of the following function prototypes (declarations)?

Code:

---

void swap( int, int );
void swap( int = 1, int = 1 );
void swap( int*, int* );
void swap( int&, int& );
void swap( int&, int* );

---

And I'm clueless. I know that if there's a void before a function, it doesn't return anything. Beyond that I don't know much.

Depends on your prof's definition of "potential problems". There are reasons to pass a straight int vs a pointer to an int vs a reference to an int, etc. etc..

Well my trouble is I just don't know what exactly those void swap( ); thingy does. If used correctly, what is it supposed to do?

You pick one to define the function a certain way. The first one takes two ints. The rest take something else.... either references or pointers. It all depends upon what you want to do.

The potential problem between the first two is that one has default parameters and the other doesn't, which can be misleading in various situations. (It wouldn't be allowed to give both of the default parameters, so the one without should be removed.)

The potential of the first two in general is that they'll have a hard time swapping their values, given that they take them by value.

The third one works.

The fourth one works, too, but is hidden by the first two if you pass variables.

The last one probably works, too, but it completely unintuitive.

Is there a way I can sort 3 integers using only the ternary operator and the usual <, <=, >, >=, ==, != ? My prof wants it for 4 integers, but I can't seem to figure out how to compare 3 digits with just the ternary operator and the above comparisons (I'm not supposed to use if else at all). Any help (I'm hoping that if I see a 3 integer comparison, I can figure out a 4 integer comparison)? It seems like I can't compare more than 2 values in one ternary operator. For example, the following code doesn't seem to work:

Code:

---

(a > b > c > d) ? (cout<<a <<b <<c <<d) : (b > c > d > a)? (cout<< b <<c <<d <<a) : (c > d > a > b) ? (cout<<c <<d <<a <<b) : (d > a > b > c)? (cout <<d <<a <<b <<c) : (cout << "=[" <<endl);

---

Does this mean that I can't embed one ternary operator into another? What are its restrictions? I'm totally lost here...

sorry for messing up the page =[

>> Does this mean that I can't embed one ternary operator into another?
You can, but you have to have the cout outside of it.

Quote:

---

(a > b > c > d)

---

This doesn't do what you think it does. A single operand cannot be operand to more than one operator.

Quote:

---

Originally Posted by CornedBee

This doesn't do what you think it does. A single operand cannot be operand to more than one operator.

---

OH!

should I have to write it like so:

Code:

---

(a>b),(c>d)

---

@Daved: You mean cout outside of the ternary operator? Can you give me an example? :(

Quote:

---

Originally Posted by Bash

OH!

should I have to write it like so:

Code:

---

(a>b),(c>d)

---

@Daved: You mean cout outside of the ternary operator? Can you give me an example? :(

---

Something like this:

Code:

---

cout << (a > b)?b:a;

---

and (a >b), (c < d) doesn't quite do what you want, unless you think that comparing (a>b), throwing away that result and using the result of (c < d) is what you want - and I don't think that's what you expected it to do. You need to use && or || when combining comparisons.

--
Mats

a < b && b < c && c < d

I see =) Thanks for all the input.

I've been reading about pointers, and there's something that has been bothering me and driving me nuts. What exactly does a pointer point to? To the address in the memory block, or to the value inside the address in the memory block? And I am totally lost when it comes to managing the dynamic memory space. I've been reading online and I start off fine, but when I get to where it explains what it points to, I get really lost. Anything I can do to make me understand what exactly is going on? I tried editing a sample code and fiddling with it to check what it does but I still can't figure it out...

And what's teh difference between these?

Code:

---

my_pointer = &my_int;
*my_pointer = &my_int;
*my_pointer = my_int;

---

I have a feeling that the third one is totally erroneous, one of them gives me the value inside the address in the memory block and the other one gives the address of the memory block. But I've no idea...

Quote:

---

To the address in the memory block, or to the value inside the address in the memory block?

---

It contains the address, thus it points to the value. But pointing to is a very abstract concept anyway.

Quote:

---

And what's teh difference between these?

---

Depends on the types involved. If my_pointer is int*, that is, a pointer to int, then the first makes my_pointer point to my_int, the second is invalid, and the third sets the value of whatever my_pointer points to to the value of my_int.
If my_pointer is int**, that is, a pointer to a pointer to int (yep, that's possible), then the first and the third are invalid and the second sets the value of what my_pointer points to (which is a pointer to int, int*) to the address of my_int, i.e. it makes the pointer that my_pointer points to point to my_int.

Quote:

---

Originally Posted by Bash

I see =) Thanks for all the input.

I've been reading about pointers, and there's something that has been bothering me and driving me nuts. What exactly does a pointer point to? To the address in the memory block, or to the value inside the address in the memory block? And I am totally lost when it comes to managing the dynamic memory space. I've been reading online and I start off fine, but when I get to where it explains what it points to, I get really lost. Anything I can do to make me understand what exactly is going on? I tried editing a sample code and fiddling with it to check what it does but I still can't figure it out...

And what's teh difference between these?

Code:

---

my_pointer = &my_int;
*my_pointer = &my_int;  Error!!
*my_pointer = my_int;

---

I have a feeling that the last one is totally erroneous...

---

Only the one marked Error in red is incorrect.

The first one says that "I have a pointer (my_pointer), it has the value of the address of my_int".
The third one says, set the value pointed to by my_pointer to the value of my_int.

Briefly: Pointers have two properties: Their own value, and the value of the data they point to. The former is the variable content itself, the latter is *pointer - so content of the memory at the address that is the pointer value.

If we do (note this code WILL NOT RUN - it is to explain things):

Code:

---

int *my_pointer = (int *)1000;

*my_pointer = 7;  // 1
my_pointer = (int *)1020;
*my_pointer = 8;

---

After this code has run [if it could run!] address 1000 in the memory would have the value 7, and address 1020 would have the value 8.

[(Do not worry if you don't understand this bit, it is just in case you want to understand more - at some point you will probably understand why this is and how it works) The reason it won't run is that you can't just assume that some random memory location, like 1000, is available to write to - it most likely won't be a valid memory address, and even if it is a valid address, it may be used for something already, so you'd be overwriting some other data. In a modern system, "low" addresses like 0 and a few thousand upwards will be permamently marked as "not present" just to catch people using a 0 pointer - 0 (or NULL) pointers should not be used, as that's an indication of an invalid pointer]

--
Mats

Quote:

---

Originally Posted by CornedBee

Depends on the types involved. If my_pointer is int*, that is, a pointer to int, then the first makes my_pointer point to my_int, the second is invalid, and the third sets the value of whatever my_pointer points to to the value of my_int.

---

I know that the "*" is a dereferencing thingy, and this is how I think about it:

Quote:

---

A pointer generally points to an address. But I see an asterisk before it, which means it's not pointing to teh address anymore, but to whatever is inside the address.

---

Is that a wrong way to think about it?

And for the first one:

Quote:

---

There is teh "&" symbol, so it's the address of whatever that follows, and that's being stored inside the pointer

---

That fine too?

And out of curiosity, is there a difference in teh following:

Code:

---

&Address and Address&
*Pointer and Pointer*

---

Edit:
I think I confirmed that I'm not wrong in my way of trying to memorize how pointers point when I studied matsp's explanation. Thanks for putting a small bit of confidence and self-assurance in me :)

Edit 2: I still would like to know the difference in whatever I've written in the code box, lol. Please?

Quote:

---

Originally Posted by Bash

I know that the "*" is a dereferencing thingy, and this is how I think about it:



Is that a wrong way to think about it?

And for the first one:



That fine too?

And out of curiosity, is there a difference in teh following:

Code:

---

&Address and Address&
*Pointer and Pointer*

---

---

It's not really "Address&" and "&Address". You should look at it as:
Variable& and &Variable. And they are (sort of) related but not quite the same.

& before something means "Take the address of the data"
& after somethign means "This is a reference". A reference is "pass this as a pointer but hide it" - or another way to put it "make this an alias of the variable specified". E.g:

Code:

---

int b = 0;
int &a = b;  // a references b.

a = 7;  // b is also 7 now.

---

*Pointer is correct - Pointer * is not - it is Type *, e.g. int *, or char *, or mytype *.

* in front means "Use the next stuff as a pointer"
* after something means "This points to what's in front", e.g. an integer, a char or a data structure.

I hope this clarifies things.

--
Mats

Quote:

---

Originally Posted by matsp

It's not really "Address&" and "&Address". You should look at it as:
Variable& and &Variable. And they are (sort of) related but not quite the same.

& before something means "Take the address of the data"
& after somethign means "This is a reference". A reference is "pass this as a pointer but hide it" - or another way to put it "make this an alias of the variable specified". E.g:

Code:

---

int b = 0;
int &a = b;  // a references b.

a = 7;  // b is also 7 now.

---

I hope this clarifies things.

--
Mats

---

It does indeed clarify a lot. But I still have a couple more questions...

1) Shouldn't

Code:

---

int &a = b;  // a references b.

---

be

Code:

---

int a& = b;  // a references b.

---

2) If I swap the address of the two variables would the values inside the address be swapped too?

Quote:

---

Originally Posted by Bash

Is that a wrong way to think about it?

---

Not exactly wrong, but I'd call it misleading.
The pointer always points to the variable and its value (the distinction between the two doesn't matter at this time - it will matter when you encounter the terms "r-value" and "l-value"). It never points to the address (it contains the address - that's what makes it a pointer).
Dereferencing doesn't change the pointer, it changes the expression. Hmm ... example without pointer:

Code:

---

int x = 5;
std::cout << -x

---

"A variable holds a value. But I see a - in front of it, which means it doesn't have the value anymore, but its arithmetic negation."
See the problem with the formulation? x doesn't change just because a - is in front of it. You've changed the expression, so the value of the expression changes. But not the part of it that is x.
* works the same way. The expression's value is no longer the address the pointer holds, but the value at that address, the thing the pointer pointed to.

Quote:

---

That fine too?

---

That's fine.

Quote:

---

And out of curiosity, is there a difference in teh following:

Code:

---

&Address and Address&
*Pointer and Pointer*

---

---

Well, three of them are invalid. ;)
An address is what you get when you do &variable. Thus, &address would be & &variable, and that's invalid. (Try it. The error message will probably contain the term "r-value" or "l-value".)
You can use postfix & on types:

Code:

---

int i;
int &ri = i;

---

This creates a reference. References are similar to pointers, but they don't advertise that fact. They are used like variables. You cannot change what they point to. You cannot have references to references. They cannot be NULL. They are, in fact, alias names for the variable they reference. In the example above you can treat i and ri exactly the same. There is no detectable difference. (And the compiler will probably completely remove the reference and use i directly.)

*Pointer
Now this takes the value of the pointer (an address) and applies the dereference operator to it. This means, as we know by now, that the result is the value at the address.

Pointer*
Syntax error. If you have type*, like matsp said, you declare/define pointers.

Quote:

---

But I still have a couple more questions...

---

1) No. The & is part of the type.

2) You cannot swap the addresses of variables. The address of a variable is its identity.