Question

Coming from a typeless language (PHP) I’m a bit confused about datatypes in C. I’m facing the following strange behavior.

//First case
unsigned int a;
a = -1;
printf("a = %u", a); //Outputs a strange number, no problem here

//Second case
unsigned int a;
a = -1;
printf("a = %d", a); //Outputs -1

What I don’t understand is how a “contained” a signed value after we specifically said it’s unsigned?

How can only formatting the output in the second case do that?

Answer 1

The -1 will be represented as something like 10000000 00000000 00000000 00000001 (if int has 4 bytes as in my case). When you are outputting it with %d, the printf() interprets it as a normal int, thus it interprets the leading 1 as the sign, and prints -1. If you output it as %u, printf() knows it’s unsigned, and interprets the leading 1 as 2^32, resulting in 4294967295 as output.