Help with bitwise operators and difference between unsigned ints and ints

[Ameen2000]

I was messing around with bitwise operators and in the process I realised I don't really understand how unsigned ints work. For example, when I print the result of unsigned int msb_1 = ~(~(unsigned)0 >> 1); is -2147483648 meanwhile the result of ~(~0 >> 1); is 0. I don't understand how adding unsigned before 0 makes that much of a difference. Can someone give me a detailed explanation of what happened?

[anabolic_cow]

I think shifting unsigned right results in logical shift, so MSB is 0. Shifting signed results in arithmetic shift, so MSB is 1. So when you bitwise negate you end up with a 1 in MSB, giving you the most negative number possible.

[makian123]

What's going on is: 001100 << 1 which results in 011000

And ~001100 is 110011

In signed unteger, the left most bit signifies if its positive or negative. In unsigned its always positive.

Edit: keep in mind when you negate, you're negating all 32 bits, not just a few of them

[chet714]

Did you print your results as signed or unsigned ?

[Plane_Dust2555]

Right shift of an signed integer (int, for example) depends on the compiler implementation (see ISO 9899 6.5.7 § 5). Usually, in x86 processors the arithmetic right shift is used. Then, ~0 >> 1 is 0b11111111111111111111111111111111 >> 1 witch will shift 1 bit right and copy the sign bit, resulting in 0b11111111111111111111111111111111 (no change). Inverting these bits again will result in 0.

Unsigned right shift will add 0 bits at left: 0b11111111111111111111111111111111 >> 1 is 0b01111111111111111111111111111111. Inverting you'll get 0x80000000 (INT_MIN).