Why it's print grabage value?

Question

In this code I want to print floating point value of 9/5 which is 1.8 than why it print the garbage value. why I try to print the int value than it print 1 which is correct but why it print grabage rather than printing 1.8?

#include <stdio.h>

int main() {
      printf("%f", 9/5);
      //printf("%d", 9/5);
    
    return 0;
}

Use `-Werror`, it will help you understand the issue. – vmemmap Jun 21 '22 at 06:21 — vmemmap, Jun 21 '22 at 06:21

score 2 · Accepted Answer · answered Jun 21 '22 at 06:20

2

The type of a variadic argument (argument in the ... part of the function declaration) passed to a function depends only on what you pass to the function (in your case 9/5), not on the string argument. 9 and 5 are of type int, so their quotient is the fractional quotient converted to int by truncation towards zero, which is 1. Then, by integer promotions, int is converted to int (if you passed (short) (9/5), then by integer promotions, this would also have been converted to int).

answered Jun 21 '22 at 06:20

Kolodez

553
2
9

got it! and if I do type casting would it also worked? – Lokesh Soni Jun 21 '22 at 06:45
What do you mean? If you pass `(double) (9/5)`, then the `double` `1.0` would be passed. – Kolodez Jun 21 '22 at 06:53

Why it's print grabage value?

1 Answers1