-5

why is that when I print a result from my function, it prints out 0? I tried using a prototype, and followed an example from my teacher, but no dice. Pretty lost here. 

#define _CRT_SECURE_NO_WARNINGS
#include <stdlib.h>
#include <stdio.h>
#include <math.h>

double energie_cinetique(double m, double v) {
    double resultat;
     resultat = 1 / 2 * m * (v * v);
    return resultat;
}

int main(void) {
    double reponse;
    reponse = energie_cinetique(10, 5);
    printf("Energie pour m=10 kg et v=5 m/s: %d", reponse);
    system("pause");
    return EXIT_SUCCESS;
}
Swice
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    Integer arithmetic. –  Nov 16 '17 at 02:58
  • Maybe use `%lf` instead of `%d` in your `printf()` statement. Also it maybe a good idea to use `1.0/2.0 * m * (v * v)` to ensure floating point arithmetic. Or else `1/2` integer division is `0` and `1.0/2=0.5` or `1/2.0=0.5` or `1.0/2.0=0.5` – Sudheesh Singanamalla Nov 16 '17 at 03:00

2 Answers2

6

Integer division with 1/2 = 0 so your result will be zero.

Change the half to 0.5 and you should be golden for the math part (also, multiplication is faster so use it instead of division when possible).

%d in the printf is for int types, use %f instead for a double.

After that it should work.

Ryan Krattiger
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2

There are two errors here:

First, 1 / 2 == 0, obviously because they are integers. Change to

1.0 / 2.0 * m *(v * v);
// OR
0.5 * m * (v * v);

Second, %d is a format specifier for signed integers. Using it to print double precision FP is UB. Your compiler should have already warned about this. You should write

printf("... %lf ...", ...);
            ^~~

Then the output should be correct.

iBug
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