在C中对负数进行按位运算

Question

What is the effect of the following?

unsigned int x = -1; x >>= 2;

I'm not quite sure what the x gets set to as it's a negative number and the type is unsigned?

Answer 1

unsigned int x = -1;

sets x to UINT_MAX (section 6.3.1.3, point 2)¹. The int -1 is converted to unsigned int for the initialistaion of x . That conversion is done by adding (or subtracting, but not here) UINT_MAX +1 to the value until it is in the range 0 .. UINT_MAX , once here.

Thus x >>= 2; sets x to UINT_MAX/4 , since bitwise right shift of unsigned integers is thus specified in section 6.5.7, point 5:

The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2 ^E2 .

¹ "Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type."

Answer 2

-1应该将unsigned int设置为0xffffffff，这是该类型的最高数字（大约48亿）-如果int的位数超过32位，则该数字应高得多。

Answer 3

It is an unsigned type initialised by a signed negative value. The shifting is still performed on an unsigned value.