What happens when I assign a negative value to an unsigned int? [duplicate]

Answer 1

What is happening is that you convert the value first to unsigned int, assigning 0xffffffff to x. Then using printf("%d\n") you will convert the value back to signed int still keeping the value of 0xffffffff. Thus printing -1.

Answer 2

Use %u instead of %d in order to print unsigned values. Then you should see 0xFFFFFFFF.

Answer 3

The "%d" format is for (signed) int values. If you use it with an unsigned value, it could print something other than the actual value. Use "%u" to see the actual value, or %x to see it in hexadecimal.

In the declaration

unsigned int x = -1;

the expression -1 is of type int, and has the value -1. The initializer converts this value from int to unsigned int. The rules for signed-to-unsigned conversion say that the value is reduced modulo UINT_MAX + 1, so -1 will convert to UINT_MAX (which is probably 0xffffffff or 4294967295 if unsigned int is 32 bits).

You simply cannot assign a negative value to an object of an unsigned type. Any such value will be converted to the unsigned type before it's assigned, and the result will always be >= 0.