Final Exam Review Question, please help

Hello all,
I am currently studying for a final exam coming up very very soon.
Instead of starting new threads for each question, I am simply going to post questions here so I don't over-thread the forum.

I have failed my previous exams upto this final and I am really trying to learn as much as I can b4 the final comes. My professor is tricky!
So thank you for those who make an effort to help me, I need all the help I can get!


1) What is the largest integer that can be stored in an int variable?

A)1000
B)1000000
C)1000000000
D)2 raised to the 32nd power minus 1
E)None of the Above

I find it hard to determine the limit of an integer.
I know that C) is the most correct out of the first four options but an integer that is larger then C) can be stored as an integer.

What is the most correct way to know what the limit is? How would I know what answer he is looking for b/c there are so many different times where I thought I found a limit and and I was incorrect...

Since a computer is binary, it can store only 0s and 1s. Hence it stores data using base 2. That implies there are only two symbols, 0 and 1. But if an integer is 32 bits, that means we can have a combination of a total of 32 digits consisting of 0 and 1.
Math should tell you that there are 2^32 different combinations. So if we start from 0, the maximum number an integer can hold would be 2^32-1.
But beware that the int data type does not necessarily have to be 32 bits. The standard guarantees no such thing.

Yet another crazy question. Your professor must be retarded. Well, if it was an unsigned int, then it would be an even more retarded question.

The solution is: you can't tell. It's not stated in the limit. Although an UNSIGNED integer USUALLY has a limit of "2 raised to the 32nd power minus 1". By saying "int" variable he doesn't tell us whether he means unsigned or signed or maybe even both combined. If you don't specify unsigned/signed in C, signed is assumed (for ints, at least), but who knows in this case.

So basically: you can't -ever- tell the maximum number of types in C/C++. Though if it is added that it's stored in two's complement with 32 bits, then you can tell.

If I saw this question on a test, I would make another option F) the answers above are all retarded answers to a retarded question, and then circle that.

I would pick D and put a note. Saying F will probably not grant you any points.
Demonstrate your knowledge!

Quote:

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Originally Posted by Elysia

I would pick D and put a note. Saying F will probably not grant you any points.
Demonstrate your knowledge!

---

I am just kidding. Of course, what you said is the wisest decision, but this is not the first stupid question I have helped Matt with for his test, so I am starting to think his instructor has a habit of making such questions.

Quote:

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Originally Posted by claudiu

I am just kidding. Of course, what you said is the wisest decision, but this is not the first stupid question I have helped Matt with for his test, so I am starting to think his instructor has a habit of making such questions.

---

I am not sure why, but it seems to be fairly common that guys who don't know what they are talking about (about C programming) seem to make it into teaching positions.

Can someone help me with this short program?

I have been working on it for a while and havent really gotten any ideas on how to tackle this problem:/

I know that I need to compare the array values and as long as they are not equal to a previous array value that was summed up then I need to add it to sum....

Code:

---

Write a function that takes in an integer array called values that is guaranteed to
 ONLY store integers in between 1 and 10, inclusive and the length of the array, and return
 the number of distinct values stored in the array from index 0 to index length-1. (For example,
if the array of length 8 stored 3, 7, 7, 1, 2, 5, 2, 2, then your function should return 5, since the
distinct numbers in the array are 1, 2,  3, 5 and 7.) Fill in the prototype below.


int numDistinct(int values[], int length) {


}

---

Quote:

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Originally Posted by EVOEx

Though if it is added that it's stored in two's complement with 32 bits, then you can tell.

---

How come if I write a program with an integer variable i.

then I set i=(2^32)-1 which is 4294967295

and I try to print i, then I gives me an error and prints out the wrong value b/c integer is too large?

Even if i= 4294967295 -10 or -50 or -100 it still does not run correctly:/

Sort the array you are given in any order you wish.

Then numbers in it will look like this (on your example):

1,2,2,2,3,5,7,7.

Then just iterate through an array and:

1) save the value you are currently at

2) go to next value

3) if the two are equal increment the repetition count

4) repeat

Quote:

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Originally Posted by matthayzon89

How come if I write a program with an integer variable i.

then I set i=(2^32)-1 which is 4294967295

and I try to print i, then I gives me an error and prints out the wrong value b/c integer is too large?

Even if i= 4294967295 -10 or -50 or -100 it still does not run correctly:/

---

Try unsigned int.
(And make sure you don't use the XOR operator (^), but instead the function pow.)

claudiu: Try not to give out full solutions.
You are essentially throwing away the solution, which is basically to formulate the logic required to solve the problem.

Let's see your actual code.

This is problematic:

Code:

---

i=(2^32)-1

---

Quote:

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Originally Posted by kermit

Let's see your actual code.

This is problematic:

Code:

---

i=(2^32)-1

---

---

Code:

---

#include <stdio.h>

int main () {
       
int i;

i=4294967295; //this equals (2^32)-1

printf("%d", i);

system("PAUSE");
return 0;


}

---

Quote:

---

Originally Posted by Elysia

Try unsigned int.
(And make sure you don't use the XOR operator (^), but instead the function pow.)

claudiu: Try not to give out full solutions.
You are essentially throwing away the solution, which is basically to formulate the logic required to solve the problem.

---

Please don't discourage claudiu for trying to help me that way he does. I absolutely agree with you that its better for me to come up with the logic myself however my time is VERY VERY limited, and showing me the logic in some cases might save me time, and help my understanding.

Thank you both for trying to help:)

Quote:

---

Originally Posted by matthayzon89

Code:

---

#include <stdio.h>

int main () {
       
int i;

i=4294967295; //this equals (2^32)-1

printf("%d", i);

system("PAUSE");
return 0;


}

---

---

Try

Code:

---

#include <stdio.h>

int main ()
{
    unsigned int i = 4294967295; //this equals (2^32)-1
    printf("%u", i);
    return 0;
}

---

I believe %u is for unsigned integer, but I'm not 100% sure.

Quote:

---

Originally Posted by matthayzon89

Please don't discourage claudiu for trying to help me that way he does. I absolutely agree with you that its better for me to come up with the logic myself however my time is VERY VERY limited, and showing me the logic in some cases might save me time, and help my understanding.

Thank you both for trying to help:)

---

Sure, but I'm just saying that giving out the full solution directly is bad idea.
Giving hints or helping someone out on the way is also very helpful.

Quote:

---

Originally Posted by claudiu

Sort the array you are given in any order you wish.

Then numbers in it will look like this (on your example):

1,2,2,2,3,5,7,7.

Then just iterate through an array and:

1) save the value you are currently at

2) go to next value

3) if the two are equal increment the repetition count

4) repeat

---

Something like this?

Code:

---

int values[length];
int i, sum;

for(i=0; i< length; i++) {
  if(values[i]==values[i])
      i++;
 
  else
  sum+=sum+value[i];
}

---

Quote:

---

Originally Posted by Elysia

Code:

---

int main ()
{
    unsigned int i = 4294967295; //this equals (2^32)-1
    printf("%u", i);
    return 0;
}

---

I believe %u is for unsigned integer, but I'm not 100% sure.

---

I suppose that our prof expects us to consider signed ints for that question. We havent learned %u.... He won't test us about it.

So, if the int is signed, then the answer would be C?

Quote:

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Originally Posted by matthayzon89

Something like this?

Code:

---

int values[length];
int i, sum;

for(i=0; i< length; i++) {
  if(values[i]==values[i])
      i++;
 
  else
  sum+=sum+value[i];
}

---

---

Whoa. There's something seriously wrong here.
What do you think this does, in terms of code?

Quote:

---

Originally Posted by matthayzon89

I suppose that our prof expects us to consider signed ints for that question. We havent learned %u.... He won't test us about it.

So, if the int is signed, then the answer would be C?

---

No, it would be 2^32/2 - 1.
The actual range of the integer would be about -2^32/2 - 2^32/2-1.

Can someone please help?

Which function always reads in a single character from the keyboard?

A)getch
B)putch
C)putchar
D)scanf
E)None of the Above

Wouldnt it be scanf? it seems too easy to be true, but thats my guess....
that OR E)

scanf reads in one value at a time but you can use one scanf for more than 1 value, thats why this question is retarded.

It's not retarded because there is a deterministic answer which is 100% correct.
Why don't you look up in the documentation what each function does?

I made some adjustments to the logic. Lol, i noticed that I had a HUGE mistake in my code, idk what I was thinking:)

Is it a bit better now?

Code:

---

int values[length];
int i, sum=0;

for(i=0; i< length; i++) {
  if(values[i]==values[i])
      i++;
      continue;
  else if(values[i]!=values[i])
  sum+=sum+values[i];
}

---

btw, for "Which function always reads in a single character from the keyboard?" the answer is getch() :D

Honestly? It's even more broken.
Start by explaining what it does. Line by line.

Quote:

---

btw, for "Which function always reads in a single character from the keyboard?" the answer is getch() :D

---

Yep. Good work.

Quote:

---

Originally Posted by matthayzon89

I made some adjustments to the logic. Lol, i noticed that I had a HUGE mistake in my code, idk what I was thinking:)

Is it a bit better now?

Code:

---

int values[length];
int i, sum=0;

for(i=0; i< length; i++) {
  if(values[i]==values[i])
      i++;
      continue;
  else if(values[i]!=values[i])
  sum+=sum+values[i];
}

---

btw, for "Which function always reads in a single character from the keyboard?" the answer is getch() :D

---

I was out for a while. What question is this in reference to?

Heh. Which part of the quote?
The loop is a response to your logic, and the last part is a response to another question not so high above about what function always reads one characters from the keyboard.

Code:

---

int values[length];
int i, sum=0;

for(i=0; i< length; i++) {  //loop while i<length which is the number of spots in the array
    if(values[i]!=values[i])
      sum+=values[i]; //first time through it stores the value of the first array spot and after that it adds the value of the old sum to the value of the next array spot...
  else
      i++; //an attempt to skip the array position if it is equal to another array position:/

---

honestly, im lost with this question....

That's why we're asking you to explain your code. Why did you code it like you did? Why did you think it will work? What do you think it will do?
If you cannot construct code from a given logic, then obviously there is some basic concept you haven't properly grasped, and we want to identify what and explain it you so that you can benefit from this knowledge and become a better programmer.

Quote:

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Originally Posted by Elysia

Heh. Which part of the quote?
The loop is a response to your logic, and the last part is a response to another question not so high above about what function always reads one characters from the keyboard.

---

Heh indeed. It would be nice if Matt came up with the first working sort algorithm that is O(n).

Ok, Matt let's start by figuring out how you intend to sort the file. (there are several sorting algorithms available for use, some more efficient than others, although efficiency should not be your biggest concern right now) That is why I suggest using an insertion sort (very readable and easy to remember for beginners) .

ohhhhhh I thought in your first response you said that no matter what order the numbers are the program automatically sorts it!! I didn't think that needed to be my first step... haha

okay, well we haven't learned any sort functions (if there are any).

I kind of have an idea of how to compare one position to the remainder of the array, which I guess could be useful in this case...


P.S
I edited comments into my last post with the code...

Ok, well if you haven't learnt any sorting then you are probably expected to come up with a solution without it.

How would you approach this problem in real life, without the use of a computer? Say I give you a series of playing cards, and for the sake of the actual problem I have an infinite number of any card available (as if I had 1000 decks of cards), and I give you 15 random cards from this magic pack of mine and ask you to count the repetitions of the same card value. (i.e. ace of spades and ace of clubs counts as a repetition here, the suite doesn't matter)

Describe how you would do this. I know it is hard to describe, because when you are actually doing it in real life, your brain is performing this algorithm without even knowing.

Quote:

---

Originally Posted by matthayzon89

1) What is the largest integer that can be stored in an int variable?

A)1000
B)1000000
C)1000000000
D)2 raised to the 32nd power minus 1
E)None of the Above

I find it hard to determine the limit of an integer.

---

INT_MAX in limits.h. Since that's not an option, the answer is E.


Quzah.

Quote:

---

Originally Posted by claudiu

Ok, well if you haven't learnt any sorting then you are probably expected to come up with a solution without it.

How would you approach this problem in real life, without the use of a computer? Say I give you a series of playing cards, and for the sake of the actual problem I have an infinite number of any card available (as if I had 1000 decks of cards), and I give you 15 random cards from this magic pack of mine and ask you to count the repetitions of the same card value. (i.e. ace of spades and ace of clubs counts as a repetition here, the suite doesn't matter)

Describe how you would do this. I know it is hard to describe, because when you are actually doing it in real life, your brain is performing this algorithm without even knowing.

---

I would go through the cards and after there is more than one of the same card I would put that card aside and increment the count of 'repeating card' by one (in my brain).

That's right. So in C programming you would need to:

1) grab each element of the array

2) search the whole array in a nested loop and figure out how many findings of that element you have; increase the counter each time.

3) go to the next element in the array

4) repeat

Quote:

---

Originally Posted by matthayzon89

Can someone help me with this short program?
Write a function that takes in an integer array called values that is guaranteed to
ONLY store integers in between 1 and 10, inclusive and the length of the array, and return
the number of distinct values stored in the array from index 0 to index length-1. (For example,
if the array of length 8 stored 3, 7, 7, 1, 2, 5, 2, 2, then your function should return 5, since the
distinct numbers in the array are 1, 2, 3, 5 and 7.) Fill in the prototype below.

Code:

---

int numDistinct(int values[], int length) {


}

---

---

The main thing here you should note is that the integers are only within 1-10 range. This means you can use a trick to solve this problem:

1. Create a NEW array of size 11. Set all members to 0
2. Loop through the given array, use the value of each element as the index for your NEW array and increment the new array at position "given array value".
3. You now have a new array with index 1-10 (your range of input characters) and the values of each element are the count that those index (values from given array) appeared in the original array.

Let me know if that is not clear enough...I'm not a writer.

EDIT: Just to note this is O(n) solution where as the nested loop idea would be O(n^2) at the best.

Quote:

---

Originally Posted by nonpuz

The main thing here you should note is that the integers are only within 1-10 range. This means you can use a trick to solve this problem:

1. Create a NEW array of size 11. Set all members to 0
2. Loop through the given array, use the value of each element as the index for your NEW array and increment the new array at position "given array value".
3. You now have a new array with index 1-10 (your range of input characters) and the values of each element are the count that those index (values from given array) appeared in the original array.

Let me know if that is not clear enough...I'm not a writer.

EDIT: Just to note this is O(n) solution where as the nested loop idea would be O(n^2) at the best.

---

Good solution! What I had suggested was more general but less efficient. Since he is taking an exam on all of this I thought it would be better if he knew how to solve this for any given set of integers.

Quote:

---

Originally Posted by matthayzon89

Can someone please help?

Which function always reads in a single character from the keyboard?

A)getch
B)putch
C)putchar
D)scanf
E)None of the Above

Wouldnt it be scanf? it seems too easy to be true, but thats my guess....
that OR E)

scanf reads in one value at a time but you can use one scanf for more than 1 value, thats why this question is retarded.

---

The only correct answer here is E, in my opinion. I don't think getch is in the C standard and thus it's not portable. And hence, on some systems getch wont even exist. Furthermore, how would a function in an embedded device without keyboard be able to read from the keyboard?
I think he intends the students to answer A. But so far, I suggest you to answer every question of the exam with "None of the above". I bet you'll be able to find errors in so many of his questions that you'll still easily pass.

Nice catch! getchar would be the correct; so again, E as mentioned above.


Quzah.

Quote:

---

Originally Posted by quzah

Nice catch! getchar would be the correct; so again, E as mentioned above.


Quzah.

---

Actually, getchar reads from STDIN, which might not be keyboard at all, but it might be a pipe or a file or whatnot.

The actual answer is: there is no such function, as a keyboard is not even required to run a C program.

Really, make the professor read this thread. Let his superiors read this thread. Maybe they realize they can't let someone teach who knows so little about what he teaches.


Although come to think of it, you may not have given us all information asking the questions. Maybe in the beginning of the exam it is stated that all questions regard Linux on an i386 processor, or whatever. Or maybe Windows 32 bits. In that case those questions all DO have valid answers.
But I'd consider it fairly bad if an entire exam is based on a single architecture.

You have to draw the line somewhere, as well. Plunging students into these depths directly may not be such a good idea.
Just as you teach lower mathematics with some lies first, you expand upon those later. The professor is probably trying to see if the student knows what has been brought up in the course without delving too deep.

Quote:

---

Originally Posted by Elysia

You have to draw the line somewhere, as well. Plunging students into these depths directly may not be such a good idea.
Just as you teach lower mathematics with some lies first, you expand upon those later. The professor is probably trying to see if the student knows what has been brought up in the course without delving too deep.

---

Yes, but by adding answers that are actually valid but won't be counted valid as it's not what the course taught you? That's fairly annoying for those who actually know what they're talking about.
I've been there... It's horrible... Trust me...

Quote:

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Originally Posted by EVOEx

Yes, but by adding answers that are actually valid but won't be counted valid as it's not what the course taught you? That's fairly annoying for those who actually know what they're talking about.
I've been there... It's horrible... Trust me...

---

On your version of the test, you choose E) and scratch out the word "keyboard" and write "stdin" above it. Also write a note on getch() "this is the best choice when using a specialized environment, but it is a non-standard function, getchar() would be a better replacement."

Problem solved.

Quote:

---

Originally Posted by EVOEx

Yes, but by adding answers that are actually valid but won't be counted valid as it's not what the course taught you? That's fairly annoying for those who actually know what they're talking about.
I've been there... It's horrible... Trust me...

---

Take a look at some old math tests and look for troubles where you have to take the root out of a negative number.
The answer to that would probably be "no solution" or "no real solution". The former is a valid answer and I'm not surprised to see, but of course it's wrong. The latter is a little more correct, but they probably have no idea what it means.
Horrible, isn't it? Yet, it's necessary.

Quote:

---

Originally Posted by Elysia

Take a look at some old math tests and look for troubles where you have to take the root out of a negative number.
The answer to that would probably be "no solution" or "no real solution". The former is a valid answer and I'm not surprised to see, but of course it's wrong. The latter is a little more correct, but they probably have no idea what it means.
Horrible, isn't it? Yet, it's necessary.

---

@nonpuz: Unfortunately, that's not always possible, especially when you get only a small sheet for only the answers, where you can only cross several options, and the rest is likely ignored completely.

Elysia: Well, I don't think it's necessary. I think it's an obligation to inform us first that at this time "we'll only be using real numbers. You probably don't know what this is at this time, though."
And I've actually gained a fair share of bonus points for using stuff I shouldn't have known yet. I've even received a "110%" grade at one point, though it had to be rounded down to 100% because they weren't allowed to give grades higher than that.

So also, in this case, you should get bonus points for writing that "getch" isn't available everywhere, in my opinion. Though I expect you'd just lose points.
Also, what's so difficult about telling the students "well, this isn't standard, but it's available on Linux and Windows and most other architectures. See, you can also code for embedded devices that need not have this function.". They'll understand.

Quote:

---

Originally Posted by EVOEx

Elysia: Well, I don't think it's necessary. I think it's an obligation to inform us first that at this time "we'll only be using real numbers. You probably don't know what this is at this time, though."
And I've actually gained a fair share of bonus points for using stuff I shouldn't have known yet. I've even received a "110%" grade at one point, though it had to be rounded down to 100% because they weren't allowed to give grades higher than that.

---

It's just one thing, though. There are lots of other things too. Like the "vanilla" coordinate system. There isn't only one.
You don't need to teach them these things. It will probably just confuse them or they will ignore it. Therefore, it is necessary to simplify some stuff.
Using stuff you aren't supposed to know yet is another thing. There's nothing wrong with that.

Quote:

---

So also, in this case, you should get bonus points for writing that "getch" isn't available everywhere, in my opinion. Though I expect you'd just lose points.
Also, what's so difficult about telling the students "well, this isn't standard, but it's available on Linux and Windows and most other architectures. See, you can also code for embedded devices that need not have this function.". They'll understand.

---

I agree that writing getch in the first place was a bad idea.
But trying to teach people about portability opens up a whole new can of worms. Suffice to say, stick to C library functions and they don't need to bring up the whole portability concept.

NOTE: that my course is Intro to C, so any complex thought about these problems above beginner level I guess my professor does not expect us to know.

There is another tricky one that I have no clue about:/

3) Which of the following is logically equivalent to the Boolean expression

(val != 0 || val != 1) ?

a) val >= 0 b) val <= 2 c) 0 d) 1 e) none of the above


What the helll? I know that 0=false and 1-true but if its not equal to zero or one? e) none of the above? or c) zero b/c its always going to be false, is since its always suppose to be 0 or 1?

To answer this, try a few values, e.g., 0, 1 and 2. It is easy to observe that:
(0 != 0 || 0 != 1) is equivalent to (0 || 1), which is equivalent to 1.
(1 != 0 || 1 != 1) is equivalent to (1 || 0), which is equivalent to 1.
(2 != 0 || 2 != 1) is equivalent to (1 || 1), which is equivalent to 1.

In fact, from just these three, it should begin to dawn on you that the expression always evaluates to 1, since the result would be the same as for val=2 for any other value. From there, a quick look at the options is enough for you to select the correct answer.

Quote:

---

Originally Posted by laserlight

To answer this, try a few values, e.g., 0, 1 and 2. It is easy to observe that:
(0 != 0 || 0 != 1) is equivalent to (0 || 1), which is equivalent to 1.
(1 != 0 || 1 != 1) is equivalent to (1 || 0), which is equivalent to 1.
(2 != 0 || 2 != 1) is equivalent to (1 || 1), which is equivalent to 1.

In fact, from just these three, it should begin to dawn on you that the expression always evaluates to 1, since the result would be the same as for val=2 for any other value. From there, a quick look at the options is enough for you to select the correct answer.

---

Oh, wow! Idk why but the notation of the problem is different than I have ever seen, in my mind I read it as something that basically always contradicts itself (ex: true and false at the same time).

This clarifies though, thank you:)
I get it, I can just come up with my own values and see what happens basically.

thanks.

Quote:

---

Originally Posted by EVOEx

Yet another crazy question. Your professor must be retarded. Well, if it was an unsigned int, then it would be an even more retarded question.

---

Given that the kid failed ALL his previous tests, what makes you think it is capable of correctly stating the given answer to this question even if handed to it on a golden platter?

Quote:

---

Originally Posted by jwenting

Given that the kid failed ALL his previous tests, what makes you think it is capable of correctly stating the given answer to this question even if handed to it on a golden platter?

---

People learn new things retard. What I didn't know then, and what tricked me then, I know now, and won't trick me now. Did you ever think of that as a possibility? Or I guess your too negative to be optimistic...



btw, while programming might be your thing, obviously English is not. Maybe, you should learn that a person is a HE or a SHE not an IT.

Insertion sort is a faster sorter, but I found Selection sort to be the easiest to memorize, by far:

Code:

---

for(i = 0; i < MAX; i++) {        //outer loop
  for(j = i+1; j < MAX; j++) {  //inner loop starts at i + 1
    if(A[i] > A[J]) {        //the comparison - change > to < for descending order sort
      temp = A[i]];        //put higher value into temp 
      A[i] = A[j];          //put lower value in A[i]
      A[j] = temp;        //put higher value in A[j]
    }
  }
}

---

And Bob's your Uncle

Good luck on your test, Matt!

I found Bubble sort to be the easiest, seeing as it was the first algorithm I pictured in my mind before I learned about all sorting algorithms out there!

Hello,
I am simply trying to run a few different functions, and mess around with the code so I get a better understand of it works. However, my code compiles but I am unable to run it. In other words, system("PAUSE") does not pause, can someone help me run this program?

Code:

---

#include <stdio.h>
int g(int *a, int *b);

int main() {
    a = 6, b = 5;
    a = g(&b, &a);
    printf("a=%d\n", a);
    printf("b=%d\n", b);
  system("PAUSE");
    return 0;
}

int g(int *a, int *b) {
    *a = *a + *b;
    *b = (*a)*3 - (*b);
    printf("a=%d\n", *a);
    printf("b=%d\n", *b);
    return *a - *b;
}

---

Quote:

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Originally Posted by Elysia

I found Bubble sort to be the easiest, seeing as it was the first algorithm I pictured in my mind before I learned about all sorting algorithms out there!

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I had the same experience.

Quote:

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Originally Posted by matthayzon89

However, my code compiles but I am unable to run it.

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Eh, your code should not compile, and indeed it does not.

Quote:

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Originally Posted by matthayzon89

However, my code compiles

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Evidently it is not the code you posted then.

Code:

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#include <stdio.h>
int g(int *a, int *b);

int main() {
   a = 6, b = 5;

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Red line is a compiler ERROR.

Dammit, why would that be on the test??

Its a question on the sample test, I am suppose to know whats going to print out....

It's probably a typo, OR a trick question....


The first code I posted DID compile but did not run... then I edited it, I apologize.

When talking about Arrays, If I am asked a question like:

"Which of the following defines an int array of size 20?"

I would say int x[19] <---- right? since 0 is a position as well?

Also, suppose I declared an an array of size 3... if I wanted to initialize all the values in all the positions to a certain number (Lets say 1), would I have to write it out like this:

int x[0]=1, x[1]=1, x[2]=1, x[3]=1;

Or is there a shortcut to achieve this? (Without loops).

Somewhere along the way the "int" went missing at the beginning of that line. Put it back, and it will compile.

Quote:

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Originally Posted by matthayzon89

When talking about Arrays, If I am asked a question like:

"Which of the following defines an int array of size 20?"

I would say int x[19] <---- right? since 0 is a position as well?

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No, int x[20]; would be correct. You are right to say that 0 is a valid index, but 19 would be an invalid index with what you proposed. Yet, it is easier to just think of int x[N]; as declaring an int array of size N.

Yes, like laserlight said int x[10] allocates 10 ints on indexes 0-9 inclusive.

Okay, thanks for clarifying! Got it;)

39) If a pointer isn't pointing to anything, to what should it be set?

a) NULL b) nada c) nothing d) X e) none of the above


NULL or E)?

btw, I hate the trickiness of his questions.

Quote:

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Originally Posted by matthayzon89

Also, suppose I declared an an array of size 3... if I wanted to initialize all the values in all the positions to a certain number (Lets say 1), would I have to write it out like this:

int x[0]=1, x[1]=1, x[2]=1, x[3]=1;

Or is there a shortcut to achieve this? (Without loops).

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Element 3 is not valid in this case.
As for the question, yes: memset.

Quote:

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Originally Posted by matthayzon89

39) If a pointer isn't pointing to anything, to what should it be set?

a) NULL b) nada c) nothing d) X e) none of the above


NULL or E)?

btw, I hate the trickiness of his questions.

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NULL, obviously.

Well I assume he wants you to answer a) however I don't think he really understands this question either.

Quote:

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Originally Posted by matthayzon89

Also, suppose I declared an an array of size 3... if I wanted to initialize all the values in all the positions to a certain number (Lets say 1), would I have to write it out like this:

int x[0]=1, x[1]=1, x[2]=1, x[3]=1;

Or is there a shortcut to achieve this? (Without loops).

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When you declare a variable, you can provide an initializer for that variable at the time that it is declared. Arrays use the curly-brace initializers.

Quote:

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Originally Posted by matthayzon89

39) If a pointer isn't pointing to anything, to what should it be set?

a) NULL b) nada c) nothing d) X e) none of the above

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NULL is a "best practice", and I would guess based on the other choices this is a heavy handed endorsement of that practice.

Setting a pointer NULL can help with memory management and prevent double frees (which crash the program):

Code:

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/* free stuff */
if (ptr) {
  free(ptr);
  ptr = NULL;
}

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This way, you know if the pointer is not NULL (the if), then it is allocated and can be freed. If it is NULL, then don't free it, or you will crash the program.

Quote:

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Originally Posted by MK27

NULL is a "best practice", and I would guess based on the other choices this is a heavy handed endorsement of that practice.

Setting a pointer NULL can help with memory management and prevent double frees (which crash the program):

Code:

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/* free stuff */
if (ptr) {
  free(ptr);
  ptr = NULL;
}

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This way, you know if the pointer is not NULL (the if), then it is allocated and can be freed. If it is NULL, then don't free it, or you will crash the program.

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The "if (ptr)" is useless there, and probably only makes the code run slower ;-). Better leave it out. As free(NULL); won't do anything.

Quote:

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Originally Posted by EVOEx

The "if (ptr)" is useless there, and probably only makes the code run slower ;-). Better leave it out. As free(NULL); won't do anything.

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I see your point, but this hinges on the contention that free(NULL) is a total no-op, as opposed to a safe and useless set of pointless operations, which I would guess the latter since this is a runtime event. In that case, you might as well skip it with the if.

I'm gonna guess that if(NULL) and free(NULL) would be nearly identical.

Thanks for all the help!
I understand that array size 3 was a mistake lol, its suppose to be array size 4.

and, if I wanted to initialize all the values in an array to 5, then I would write it like this?

Code:

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int x[20]={5};

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Below is a short response question the involves a function, I modified the code so I can test, so there IS no function, but it should still work, however, my program does not compile and when it does it does not result in the write answer.

I was trying to use his example and come up with the same output.

I want everyone to know that I really appreciate all the help up until now! ...Before I would never be able to understand how to think about the logic of how to approach such a problem, but I am finally starting to get it!

Can someone please help me understand why my program is not completing the task?

Code:

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/*3) (10 pts) Write a function that takes in as input an array of integers,
the length of the array, and an integer k as parameters and returns the number
 of integers stored in the array that are less than k. (For example, if an array
  of length 6 passed to the function stores 3, 1, 8, 5, 5, and 6 and k = 5 is passed
  to the function, the function should return 2, since only 3 and 1 are less than 5.)*/

#include <stdio.h>

int main () {
       
int arr[6];
int sum, i=0, k=5, length=6;
arr[0]=3, arr[1]=1, arr[2]=8, arr[3]=5, arr[4]=5, arr[5]=6;

while(i < length)
  if(arr[i] < k)
  i++;       
 

printf("%d\n", i);
       
       
system ("PAUSE");
return 0;

}

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Again, because of your flaw in logic.
"i" in this case is a variable that you use to keep track of what element you are comparing to.
You need to keep track of how many matches you've found.

Newer code, still not working.

...I noticed that printing 'i' was a mistake, second too late. Thanks.

Code:

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#include <stdio.h>

int main () {
       
int arr[6];
int sum, i=0, k=5, length=6;
arr[0]=3, arr[1]=1, arr[2]=8, arr[3]=5, arr[4]=5, arr[5]=6;

while(i < length){
 
  if(arr[i] < k)
  sum+=i;
  i++;
}

printf("%d\n", sum);       
       


       
system ("PAUSE");
return 0;

}

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sum is not initialized.
You need curly braces around your while if it's more than one statement long.
And why the heck do you add i to sum when you find a match?

Code:

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int arr[6];
int sum=1, i=0, k=5, length=6;
arr[0]=3, arr[1]=1, arr[2]=8, arr[3]=5, arr[4]=5, arr[5]=6;
while(i < length){
 
  if(arr[i] < k)
  sum+=i;
  i++;
 
}
printf("%d\n", sum);

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Got it!

I add sum to 'i' b/c if you read the instructions for the problem, I need to keep count of how many times there are numbers less then k, and return that 'count' to the main (as if the code above was a function)


Thanks for your help.

Although my logic seems wayy off at times, I have come a long way, and am able to handle these types of problems better then I used to before.

Quote:

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Originally Posted by matthayzon89

I add sum to 'i' b/c if you read the instructions for the problem, I need to keep count of how many times there are numbers less then k, and return that 'count' to the main (as if the code above was a function)

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Your logic isn't quite correct.
Before starting, you haven't found "any" numbers, have you? So sum should be initialized to 0. But then you get the wrong output. Why? Because you add i to sum instead of the right thing.
So the first loop, arr[i] < k, so we add 0 to sum (makes no sense, does it, since we found a number less than k and we should count the number of occurances?).
Second loop, we find arr[i] < k, so we add 1 to sum. Again, makes no sense.
All other numbers in the array are >= k, so no problems here. But what if the last element in the array, and only the last element in the array was < k? Then you would add 5 to sum. That makes no sense whatsoever, does it? Since you would have found only 1 number < k in that case.

You are wrong. My logic is in fact correct. I initialize sum at 1 b/c there is an 'if statement' after the loop. If the conditions are ever met at all then there is automatically 1 added to the count. Every time after that, that the conditions in the 'if statement' are met, sum is increased by 1.

If sum was initialized to zero and the if statement conditions were met then then sum would still equal zero b/c i=0 and sum=0.

What I should do is actually add this to my code:

Code:

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else
sum=0;

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in the case that none of the array values satisfy the if statement condition.

Let's try that, shall we?
Test code:

Code:

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#include "stdafx.h"
#include <stdio.h>

int ReturnLessThan(int arr[6], int length, int k)
{
        int sum=1, i=0;
        while(i < length)
        {
            if(arr[i] < k)
                  sum+=i;
          i++;
          }
        printf("%d\n", sum);
        return sum;
}

int main()
{
        int arr[] = { 1, 2, 3, 4, 5, 6 };
        ReturnLessThan(arr, sizeof(arr) / sizeof(int), 5);
}

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Basically, ReturnLessThan is just your algorithm copied over. I just added a return sum.
Now, according to the spec, this should print out 4, because there are 4 numbers that are less than k, right?
What does it print out, then? It prints 7. That's not right, is it?

If we change arr to

Code:

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int arr[] = { 5, 6, 7, 8, 9, 1 };

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It gets even more drastic and prints out 6, instead of 1.

I think I get what your saying, you have a point!

Let me work on the logic...

Is this better?
This should make more sense...

My logic worked for one condition but not for all conditions which is VERY bad in programming haha.

Here is my new code:

Code:

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int main () {
       
int arr[6];
int sum=0, i=0, k=5, length=6;
arr[0]=5, arr[1]=5, arr[2]=8, arr[3]=5, arr[4]=3, arr[5]=6;


while(i < length){
 
  if(arr[i] < k)
  sum++;
  i++; 
 
}

printf("%d", sum);

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Perfect!