What should be the value of si?

Question

I am trying to workout the output of the below given c program but unable to figure out.

Why is si = 0?

#include<stdio.h>

int main(void){

    float p = 1000;
    float r = 5;
    float t = 2;

    float si = p * r * t / 100;

    printf("SI = %d", si);

    return 0;

}

`printf("SI = %d", si);` but `si` is a `float`: change to `%f` — Jean-François Fabre, Aug 29 '17 at 18:18
It should print the 4 bytes of float si by interpreting them as an integer. If you want it to print 100 (or 99) then you can typecast `si` as `int` in the print statement. — Vikas Yadav, Aug 29 '17 at 18:22
good c beginners book needed. **this topic is the multiple duplicate** — 0___________, Aug 29 '17 at 18:29
Printf isn't that smart. You'll have to do the `(int)si` cast yourself. — Lee Daniel Crocker, Aug 29 '17 at 18:34

kocica · Answer 1 · 2017-08-29T18:26:06.800

2

You are passing value in floating point representation while you used %d specifier which forces printf to expect an int.

There is no 'automatic cast' when you pass variables to a variandic function like printf, the values are passed into the function as the datatype they actually are (or upgraded to a larger compatible type in some cases) because c is weakly typed language.

If you pass type with incompatible internal representation you will get garbage output.

Change

printf("SI = %d", si);

to

printf("SI = %f", si);

or (if you would like to get integral value of expression)

printf("SI = %d", (int)si);

edited Aug 29 '17 at 18:26

answered Aug 29 '17 at 18:19

kocica

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any idea why is he getting 0 instead of expected garbage value? – Vikas Yadav Aug 29 '17 at 18:28
1

@VikasYadav Consider: I send you a `float` and put it in the `float` box. You look in the `int` box and see ... nothing .... and print that. Perhaps a 0 left over from before, maybe not, maybe the boxes overlap? It is _undefined behavior_. – chux - Reinstate Monica Aug 29 '17 at 18:57

What should be the value of si?

1 Answers1