Negative integer bitswitching

[oseri]

I'm wondering why does the following code loops infinitely :

int temp = -4;
	while (temp)
	{
		temp >>= 1;
	}

As I understand it, this loop should go through all 32 bits that an integer consists of and then exit, but that seems not to be the case. Why ?

I get it that ints are stored in U2 system, but that shouldn't prevent me from switching through all the bits leaving only zeros.

[Disch]

Your system is doing an arithmetic shift, and not a logical shift. This is typical of signed integers, as it more closely simulates a divide-by-2 operation and is much more useful in practice.

A logical shift right uses 0 as input for the shifted in high bit.
An arithmetic shift right uses the value of the current high bit for the shifted in high bit.

IE:

8-bit signed:    11100101

>> 1 logical:    01110010
>> 1 arithmetic  11110010


For arithmetic:  11100101
                 v
                 |
                 this bit used to shift in
                 |
              +--+
              |
              +->11110010