32位系统中的无符号短整数提升

Question

I have a 32-bit integer size

if I have an arithmetic expression like

unsigned short current_time, last_time
if((current_time - last_time) > timeout)

I believe current_time and last_time will both be converted to signed int 32 before the subtraction. There are no problems with a 16-bit integer size system, but with this 32-bit integer size system will there be potential for a negative value because of the integer promotion?

Answer 1

If current_time is greater than or equal to last_time , there won't be potential for a negative value.

To quote section 6.2.1.2 "Signed and unsigned integers" of the C90 spec:

When a value with integral type is converted to another integral type, if the value can be represented by the new type, its value is unchanged.

With unsigned short being shorter than int , all values of type unsigned short can be represented by int , so the converted-to- int values of current_time and last_time will be the same as their un-converted unsigned short value, and the result of the subtraction will be what you expect it to be.

If, however, current_time is less than last_time , there is potential for a negative value; that, however, is not a bug, it's a feature, because, in that case, time really did go backwards.

If current_time and last_time were unsigned int s, and current_time were less than last_time , the result of the subtraction would be an unsigned int , and thus not negative; it would be the difference between the times, modulo the maximum value of an unsigned int .

You probably want to handle time going backwards specially, anyway, if it can happen.

Answer 2

根据现代C规则，无符号短裤将提升为有符号整数，并且差异将根据其原始值正确签名。