In C, my output of a function is always 0.000000. Is it because the two inputs are int? [duplicate]

Question

I know you're usually not meant to put out all of your code but this is short and would help with the problem. Can anyone please explain why the output is 0 and how I can change the code to output what should be the volume of a cone.

#include <stdio.h> float ConeVolume(int height, int radius); float ConeVolume(int height, int radius) { float pi; pi = 3.14159; float third; third = (1/3); float vol; vol = third * pi * radius * radius * height; return vol; } int main() { float x = ConeVolume(12,10); printf("%.4f \n", x); }

edit: thank you to all who answered so quickly. Great community here.

Compile with all warnings & debug info (gcc -Wall -Wextra -g). Then use the debugger (gdb) by single stepping your program. You would have found your bug quickly (faster than asking here). — Basile Starynkevitch
– Basile Starynkevitch, Commented Oct 4, 2015 at 11:29

alk · Accepted Answer · 2015-10-04 11:33:40Z

7

1/3

is an integer division and always results in 0.

To have this evaluate to a floating point variable you might do

1./3

or

1/3.

or

1./3.

or even more explicit

(float)1/(float)3

for example.

edited Oct 4, 2015 at 11:33

answered Oct 4, 2015 at 11:28

alk

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3 Comments

Johannes Schaub - litb Over a year ago

I was about to add (not quite serious) that it's not the ./ operator in action there :) But then you showed the other variations and I think it became clear that you are dividing floats :)

alk Over a year ago

Ah yes, missed the ./-operator ... dxxm, failed again ;-) @JohannesSchaub-litb

Basile Starynkevitch Over a year ago

It is not a ./ operator, it is a / operator applied to 1. and 3

user5152421 · Accepted Answer · 2015-10-04 11:35:46Z

Try this;

#include <stdio.h> float ConeVolume(int height, int radius) { float pi, vol; pi = 3.14159; vol = (pi * radius * radius * height) / 3; return vol; } void main() { float x = ConeVolume(12,10); printf("%.4f \n", x); system("pause"); }

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