Value of complex algebraic expression remains zero

Question

I have this code in C, and for some reason instead of properly working with numbers, it outputs zero all the time. Can someone explain me what is going on here? I know C#, but not C.

#include <stdio.h>

int main(void) {
    // I want to express 1/6n*(n + 1)(2n + 1)

    int n = 1;
    while(n != 0){
        scanf("%d", &n);
        printf("%d", 1/6 * n * (n + 1) * (2 * n + 1));
    }

    return 0;
}

Thanks in advance!

I am using Code::Blocks + GCC compiler.

Modifying it to `printf("%d", 1.0/6.0 * n * (n + 1) * (2 * n + 1));` does not help either. — Károly Ozsvárt, Jul 05 '19 at 16:40
`%d` takes an `int`, not a `double`. You need to enable compiler warnings. — melpomene, Jul 05 '19 at 16:41
In that case the result will be `double`, but you use the format string for `int` — Eugene Sh., Jul 05 '19 at 16:41
Apparently it works the same way in C#: https://stackoverflow.com/questions/10851273/why-does-integer-division-in-c-sharp-return-an-integer-and-not-a-float — melpomene, Jul 05 '19 at 16:49
@melpomene except that format specifiers are not needed there because `.ToString()` works fine. — Károly Ozsvárt, Jul 05 '19 at 16:50
Use `printf("%f", 1/6.0 * n * (n + 1) * (2 * n + 1));` - you need to make at least one of your division operands floating point, and you need to use `%f` to format the output as floating point. — John Bode, Jul 05 '19 at 19:27

Károly Ozsvárt · Accepted Answer · 2019-07-05T16:53:58.613

1

The solution is:

#include <stdio.h>

int main(void) {

    int n = 1;
    while(n != 0){
        scanf("%d", &n);
        printf("%d", n * (n + 1) * (2 * n + 1) / 6);
    }

    return 0;
}

edited Jul 05 '19 at 16:53

answered Jul 05 '19 at 16:46

Károly Ozsvárt

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Well. Not really. The *right* solution is `printf("%d", n * (n + 1) * (2 * n + 1) / 6);` – Eugene Sh. Jul 05 '19 at 16:48
What is the difference? – Károly Ozsvárt Jul 05 '19 at 16:49
Floating point is evil, so avoid it when possible. – Eugene Sh. Jul 05 '19 at 16:50
Thank you for your help! – Károly Ozsvárt Jul 05 '19 at 16:54

Value of complex algebraic expression remains zero

1 Answers1