Dividing 1/n always returns 0.0 [duplicate]

Question

I am trying to calculate p1=(1/1)*(1/2)*...*(1/n) but something is wrong and the printf gives me 0.000...0

#include <stdio.h>

int main(void) {

    int i,num;
    float p3;

    do {
        printf ("give number N>3 : \n" );
        scanf( "%d", &num );
    } while( num <= 3 );

    i = 1;
    p3 = 1;  

    do {
        p3=p3*(1/i);
        printf( "%f\n",p3 );
    } while ( i <= num );

    printf("\nP3=%f",p3);
    return 0;
}

Forgetful

The division in p3*(1/i) is integer division. Try p3 * (1.0 / i) instead. Oh ... and use double rather than float for your floating-point numbers. – Rodmun

Isn't the second loop infinite? It waits for a condition to be met on i and num, yet it does not alter either one of them. – Intact

sorry i didnt copy the entire code – Forgetful

What about doing p3 /= i; instead? ;-) – Trapp

Answer 1

(1/i)

i is an int, so that's integer division, resulting in 0 if i > 1. Use 1.0/i to get floating point division.

Humidistat

Answer 2

1 is an integer, i is an integer. So 1/i will be an integer, ie the result will be truncated. To perform floating-point division, one of the operands shall be of type float (or, better, of type double):

p3 *= 1. / i;

Toomin

and why double instead of float? – Forgetful

@Forgetful it has more precision. – Welter

Answer 3

I had the same issue. The basic case:

• when you want to get float output from two integers, you need to convert one into float

  int c = 15; int b = 8; printf("result is float %f\n", c / (float) b); // result is float 1.875000 printf("result is float %f\n", (float) c / b); // result is float 1.875000

Magnetograph