casting

Hi,

a question on castings.

My code example:

#include <stdio.h>

int main( void )
{
unsigned int a = 4294967295U;
signed int b = 4294967295U;
signed int c = (signed int) a;

printf( "a:%ud\n", a );
printf( "b:%ud\n", b );
printf( "c:%ud\n", c );

return 0;
}



The output is:
a:4294967295d
b:4294967295d
c:4294967295d

I don't understand why "b" and "c" are also a 32-bit values. Since I
defined "b" and "c" as signed int, there are ony 31 bits that can be used
to represent the number, thus the value range is [-2147483648, 2147483647].
When assigning "b" the value 4294967295U, I thought that an implicit cast
is performed that converts the value to a 31bit value + 1 bit for the sign.
In a similar way, "c = (signed int)a" is an explicit cast that should
convert the 32bit value of "a" into a 31bit value represented by "c".
However, printf indicates that casting is not performed. Why?

Regards,
Chris
.

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