Our professor asked us to make a C program that will display the cube of a number using a while loop

Question

When I input a negative integer the output will not result to it's cube value. I'm confused what should I do? here is my code:

#include <stdio.h>
int main()
{
    
    int n, i, cube;
    printf("Enter an integer: ");
    scanf("%d",&n); 
        
    while (i<=n)
    {   
        cube = i*i*i;   
        i++;

    }
    printf("The cube of %d = %d\n ", n, cube);  

    return 0;
}

`i` is uninitialized, as is `cube`. Therefore there's no guarantee `cube` will have a determinate value after the loop (because the loop body may never be executed). There's absolutely no point in having this loop in the first place. Explain why `cube = n * n * n;` is not sufficient. — paddy, Nov 08 '21 at 23:28
Maybe he wants you to print the cubes of all numbers from 1 to n. — Barmar, Nov 08 '21 at 23:29
This is the reason most courses have lab sessions with tutors on hand to assist with understanding. Does your institution not have such a learning structure? — paddy, Nov 08 '21 at 23:31
If it is a requirement that you use a loop, you will probably want to initialize `cube=1` then loop 3 times multiplying it by n (`cube *= n`). But as observed above, there is no technical need for a loop and this is equivalent to `cube=n*n*n`. — EdmCoff, Nov 08 '21 at 23:32
You could try something like `i=3; cube=1; while (i) { cube *= n; i--; }`. In your example, you should initialize `i`. And notice that, if you initialize it to zero, for example, and you keep incrementing it, it will take a while for it to be smaller than a negative `n` (it has to overflow first). — rturrado, Nov 08 '21 at 23:32

score 1 · Accepted Answer · answered Nov 08 '21 at 23:37

1

I would assume you'd want to compute the cube of a number by using a loop that iterates 3 times.

int n, cube, i;
printf("Enter an integer: ");
scanf("%d",&n); 

cube = 1;
i = 0;
while (i < 3)
{
    cube = cube * n;
    i++;
}

printf("%d\n", cube);

answered Nov 08 '21 at 23:37

selbie

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This is correct. Short, sweet, and simple. – Jaguar Nation Nov 09 '21 at 00:16
Oh, didn't know it was this simple lol. Thank you sir :) – gigibot Nov 09 '21 at 04:55

0___________ · Answer 2 · 2021-11-08T23:44:22.360

0

Loop is wrong.
Your loop invokes undefined behaviour as i is not initialized.

You simply need:

long long cube(int n)
{
    return n * n * n;
}

int main(void)
{
    int x = -21;
    while(x++ < 40)
        printf("n=%d n^3 = %lld\n", x, cube(x));

}

Or if you need to use while loop:

long long cube(int n)
{
    long long result = 1;
    int power = 3;
    while(power--) result *= n;
    return result;
    
}

Or full of whiles and no multiplication.

long long cube(int n)
{
    long long result = 0;
    unsigned tempval = abs(n);
    while(tempval--) result += n;
    tempval = abs(n);
    n = result;
    while(--tempval) result += n;
    return result;
    
}

edited Nov 08 '21 at 23:44

answered Nov 08 '21 at 23:30

0___________

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1

The question says that they're required to use a `while` loop. – Barmar Nov 08 '21 at 23:30
@Barmar amended, only while loops :) – 0___________ Nov 08 '21 at 23:36

score 0 · Answer 3 · answered Nov 09 '21 at 00:32

0

You just want the cube of 1 number?

i = 0;
cube = 1;
while (i<3)
{ 
    cube *= n;
    i++;
}

answered Nov 09 '21 at 00:32

stark

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Our professor asked us to make a C program that will display the cube of a number using a while loop

3 Answers3