Thread: Lost ("short int" and "while" statements)

You need to find out the highest number a variable of short int can store.
A variable can only hold a finite number, remember.
So the exercise asks you to use a loop to find the highest positive number it can store and the highest negative number it can store.

Big hint: if you add one to the largest value a data type can store, it "wraps" with overflow. In other words, <max>+1 is 0, for an unsigned number, or something negative for a signed number.

No way, no how. The logic is correct, but the code is incorrect.
While you're at it, indent the code and let main return int.
And you shouldn't spoil homework for others. They won't learn.

Nope.
First you can't do "x, y" in a condition and second you don't have to.
You aren't supposed to assign in the condition.

More correctly:

Code:

#include <stdio.h>
int main()
{
        short int c = 0;
        short int lowest;
        short int largest;

        do
        {
           lowest = c--;
        }  while(c < 0);

	largest = c;
	printf("Smallest: &#37;d Largest %d", lowest, largest);
}

even better, just use limits.h

Screw the while loop, that's what limits.h is for.

Code:

#include <stdio.h>
#include <limits.h>

int main(void)
{
    printf("Short min: &#37;d, Short max: %d\n", SHRT_MIN, SHRT_MAX);
    return 0;
}

It's obvious that's not what they want you to do... but the above is how I'd do it -- even if I was given this problem.

I believe the exercise is more about to find out how memory and unsigned/signed and variables works, not finding out about the maximum value they can hold.

First you can't do "x, y" in a condition and second you don't have to.

First -- you're wrong. Second, I concur.

Ever seen code like this?

Code:

for(x = 0, y = 1; ...

That's using the comma operator. Same idea anywhere else: the left side is evaluated, then the right side, and the whole expression gets the value of the right side. So this

Code:

do
    code();
while(expression());

is the same as

Code:

do
    ;
while(code(), expression());

(Your compiler might want you to put extra parentheses around there, you never know. Oh, and the semicolon is necessary. )

I was going to suggest limits.h, too. If you read the problem carefully:

Your program MUST use while statement (and break if necessary) only for looping.

In other words, when your program loops, you have to use a while loop, and not, say, a for loop. But I figured that would be temping fate . . . .

There's another way you can do this problem, much more efficiently, and still using a while loop. Consider the fact that the maximum value of any data type with most systems is one less than a power of two . . . or that an unsigned value, subtracted by one, gives you the maximum unsigned value, and the signed value should be easy to figure out from there. Okay, you don't need a loop for that solution. And both of these are rather unportable, so just ignore the suggestions -- just some random thoughts.