CODEWITHSUNDEEP
cpow
IdotEXE
IdotEXE

Reputation: 17

Why does pow output 0 even when using %f?

I am completely new to c, so I was trying to use the pow function as I'm following a tutorial. On the tutorial (and I followed the code exactly as they wrote it), their output is correct, but all I get is 0.000000. Does anyone know why this happened?

#include <stdio.h>
#include <stdlib.h>

int main() {

    printf("%f", pow(4, 3));



    return 0;
}

Output

0.000000

Upvotes: 0

Views: 428

Answers (2)

fares
fares

Reputation: 93

If you work in linux and you want to insert #include <math.h> don't forget to compile with -lm to link the math library.

And by the way if you want to use float numbers try to use powf.

Syntax: float powf(float___x,float___y);.

Upvotes: 1

Eric Postpischil
Eric Postpischil

Reputation: 223795

Insert #include <math.h>, turn on warnings in your compiler, pay attention to warning messages, and promote warnings to errors.

Without <math.h>, pow is not declared, and pow(4, 3) passes arguments as int. pow needs its arguments passed as double, the behavior when they are passed as int is not defined by the C standard. Further, the return type of pow will be assumed to be int, but the actual return type is double, and the behavior of the function call with this mismatch is also not defined. And passing an int value for a printf conversion of %f also has behavior not defined by the C standard.

Upvotes: 8

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