do and while loops

Hi this is my first post, I am having a problem with loops. I am reading through South-Western's Introduction to C++ 3rd ed. In chapeter 6, activity 6-2 it tells you to make a program that asks or a series of integers one at a time except for 0. When you enter 0 the program takes the numbers you put in and finds the average, the largest number you put in, the smallest number you put in, and the range of the numbers you put in. I got it to do all of the following, but for some reason it wont print to the screen. I think I just have my statements in the wrong order or something, I'm using Dev-C++ and here is my source:

Code:

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#include <iostream>
using namespace std;

int main()
{
    int i, lar, sma, non;
    float av;
    av=0.0;
    non=0;
    sma=2147483647;
    lar=-2147483648;
    do
    {
          cin >> i;
          cout << "You Entered " << i << endl;
          av=av+i;
          non+=1;
         
          if(i>lar)
          {lar=i;}k
         
          if(i<sma)
          {sma=i;}
         
          if(i=0)
          {cout << "Average = " << av/non << endl;
          cout << "Largest = " << lar << endl;
          cout << "Smallest = " << sma << endl;
          cout << "Range = " << lar-sma << endl;
          break;}
    }
    while(1);

return 0;
}

---

Can anyone help me out? I've been messing around with it for an hour and decided to ask for help.

after the first if statement you have a 'k' all by itself, must be a typo, get rid of it. it shouldnt compile with this in it anyway.

your third if statement (if (i=0)) is the solution to your problem. what is the '=' (assignment) operator for? what should you be using instead? if you dont know search for c++ operators..

also, besides having this last if statement, you should change your while(1) to loop while 'i' does NOT equal '0' (zero) then move the 4 couts and put them after the while statement, and get rid of the break.... this is why do loops have a while clause, for the purpose you are trying to achieve, that is: "do something while [condition]". in your case, 'do something while i does not equal zero'.

if you dont understand try and edit as much of your code as i have said and post it.. or if you get other errors post the updated code and error message.

good luck.

if anyone else is watching, how come sma and lar can store the values he has assigned to it? is the default 'int' = 'long int'?

Quote:

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Originally Posted by nadroj

if anyone else is watching, how come sma and lar can store the values he has assigned to it? is the default 'int' = 'long int'?

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An int is a 32 bit value, so those numbers are perfectly valid and are the min/max signed values of a 32 bit integer.

Also, to cj_h4x0r:

Code:

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if(i=0)

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As stated by nadroj, this line will cause the problem. Think of the code this way:

Code:

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int i = 10;
i = 0;
cout << i << endl;

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What is i at the end of this program? You should be saying 0.

Applying this same logic to the if statement, you are assigning i to become 0, so you then are testing if (0) which will always be false. This is a very common error when beginning programming, so don't fret. Merely switch the = (assignment) operator to the == (comparison) operator and you will be fine.

i thought the max for int was (-/+)32XXX? ya i did sizeof on both int and long and they are both 4 bytes.. why can long int store such a greater value while occupying the same amount of memory.

Uh ...

All types are of course largely platform-dependent, but on the x86 sizes are like thus:
char: 8 bits. Unsigned 0 - 255, signed -128 - 127.
short: 16 bits. Unsigned 0 - 65xxx (2^16), signed -32xxx - 32xxx.
int == long: 32 bits. Unsigned 0 - 4 billion something, signed -2 billion - 2 billion.

The values of sma and lar happen to be INT_MIN and INT_MAX of a 32-bit signed integer.

if int and long can store the same range whats the difference?

They do not necessarily have the same range, but it happens to be the case here.

due to what? i know it works on my system too (VS.net 8).

ive read afew sites that say the ranges as i had said.... but just a second ago i say one that says its for 'portability'. ie, an 'int' can store (no matter what OS or CPU) -/+32k, while long int can no matter what store -/+(whatever the number)

in most cases long int is the same thing as int. The same range and everything.

> ie, an 'int' can store (no matter what OS or CPU) -/+32k
The standard expresses such things as minimum requirements, not absolute requirements.

That is, and int needs to be able to store at least +/-32K, not that it has to be limited to only +/-32K.

However, where the standard allows the freedom to choose, most implementations go with the default machine 'word' for storing ints where this exceeds the minimum standard requirements.

An int is only guaranteed to hold from -16383 to +16383.

> An int is only guaranteed to hold from -16383 to +16383.
Read Dave's links here - http://cboard.cprogramming.com/showthread.php?t=84349
Annex E of the first one is C99 TC2, which states
#define INT_MAX +32767
#define INT_MIN -32767

The C89/C90 versions say the same.

Quote:

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Originally Posted by nadroj

due to what?

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Due to your compiler manufacturer's choices. If you don't like it, write your own ;)

On my Linux AMD64 box, long is 64 bits while int is still 32. But only under GCC. If I were to install a 64-bit Windows and run a 64-bit VC++ there, its compiler would still use 32 bits for long.