On 06/02/18 16:22, Liu Hao wrote:
> On 2018/2/6 23:03, Peter Breuer wrote:
>> int main() {
>> signed int x = 0x80000005u;
>> unsigned int y = 0x00000002u;
>> signed int z = x >> y;
>> printf("0x%0x\n", z);
>> return 0;
>> }
>> % gcc -std=c99 test.c
>> test.c: In function 'main':
>> test.c:6:3: warning: implicit declaration of function 'printf' [-Wimplicit-function-declaration]
>> printf("0x%0x\n", z);
>> ^
>> test.c:6:3: warning: incompatible implicit declaration of built-in function 'printf'
>>
>> (I'll live with that warning - just for the purpose of a clean example!)
>>
>
> Well you could have declared the `printf()` function:
>
> extern int printf(const char *restrict format, ...);
Or, far better :
#include <stdio.h>
<snip>
> Clearly, the type of the operand on the right-hand side is not altered -
> it remains `unsigned int`.
>
Equally relevantly, the left-hand side is not altered and remains
"signed int".
Basically, for operators that are at least roughly symmetrical (like *,
+, /, -, etc.), arithmetic conversions are used to make both sides into
the same types (of at least "int" size). But for asymmetric operators,
like <<, >>, ?:, the comma operator, you get the integer promotions but
they are handled separately.
It all makes sense when you think about it.
