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Home/ Questions/Q 7910929
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T13:03:32+00:00 2026-06-03T13:03:32+00:00

I have this simple program #include <stdio.h> int main(void) { unsigned int a =

  • 0

I have this simple program

#include <stdio.h>
int main(void)
{
 unsigned int a = 0x120;
 float b = 1.2;
 printf("%X %X\n", b, a);
 return 0;
}

I expected the output to be

some-value 120  (some-value will depend on the bit pattern of `float b` )

But I see

40000000 3FF33333

Why is the value of a getting screwed up? %X treats its arguments as signed int and hence it should have retrieved 4 bytes from the stack and printed the calue of b and then fetching the next 4 bytes print the value of a which is 0x120

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-03T13:03:33+00:00Added an answer on June 3, 2026 at 1:03 pm

    Firstly, it’s undefined behaviour to pass arguments to printf not matching the format specifiers.

    Secondly, the float is promoted to double when passed to printf, so it’s eight bytes instead of four. Which bytes get interpreted as the two unsigned values expected by the printf format depends on the order in which the arguments are pushed.

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