How this printf works in this program -clarification in c?

Question

Need some clarification...

Why do I get 2.50 0 0 0.0 as output?

#include<stdio.h> int main() { float a=5.0,b=2.0; printf("%f %d\n",a/b,a/b); printf("%d %f",a/b,a/b); return 0; }

The mismatched parameters in your printf statements will result in Undefined Behavio(u)r - there is nothing further to explain. — Paul R
– Paul R, Commented Mar 19, 2012 at 21:42
@Mysticial: Your edit changed the meaning of the question (though both versions have undefined behavior). — Keith Thompson
– Keith Thompson, Commented Mar 19, 2012 at 21:57
@KeithThompson I think my initial edit crossed with the OP's. I'm 100% sure it was originally int. I'll fix that then. — Mysticial
– Mysticial, Commented Mar 19, 2012 at 22:01

Kerrek SB · Accepted Answer · 2012-03-19 22:03:48Z

You are causing undefined behaviour, since the type of a/b is (promoted to) double, which does not match the format specifier %d (which expects an int).

(The reason you see 0 is probably because the sizeof(int) bytes you happen to be accessing are all zero, being part of the (very short) mantissa of a simple number like 2.5, and your platform stores floating point numbers as IEEE754 in little endian order:

 | <-- * --> // * = sizeof(int) 400 | 4 0000 0000 0000 // == 2.5 S+E | Mantissa

Try 2./5. to see some other results.)

@CarlNorum: Hm, the question got edited... Now I don't know but the OP's described behaviour suggests that a/b gets passed as a double (at least after the variadic standard promotion).

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