c - 使用 gcc 进行位移位的意外行为

Question

我有一个这样的测试程序:

int main()
{
    unsigned n = 32;

    printf("ans << 32 = 0x%X\n", (~0x0U) << 32);
    printf("ans >> 32 = 0x%X\n", (~0x0U) >> 32);

    printf("ans << n(32) = 0x%X\n", (~0x0U) << n);
    printf("ans >> n(32) = 0x%X\n", (~0x0U) >> n);

    return 0;
}  

它产生以下输出:

ans << 32 = 0x0  ... (1)  
ans >> 32 = 0x0  ... (2)  
ans << n(32) = 0xFFFFFFFF  ... (3)  
ans >> n(32) = 0xFFFFFFFF  ... (4)   

我期望 (1) 和 (3) 相同，以及 (2) 和 (4) 相同。

使用gcc版本:gcc.real (Ubuntu 4.4.1-4ubuntu9) 4.4.1

发生了什么事？

Answer 1

根据 C standard，按类型大小移动是未定义的行为, § 6.5.7.3:

6.5.7 Bitwise shift operators
(...) If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior is undefined.

你的编译器应该警告你:

$ gcc shift.c -o shift -Wall
shift.c: In function ‘main’:
shift.c:5:5: warning: left shift count >= width of type [enabled by default]
shift.c:6:5: warning: right shift count >= width of type [enabled by default]

如果您查看 assembler code gcc 正在生成，您会看到它实际上是在编译时计算前两个结果。简化:

main:
    movl    $0, %esi
    call    printf

    movl    $0, %esi
    call    printf

    movl    -4(%rbp), %ecx  ; -4(%rbp) is n
    movl    $-1, %esi
    sall    %cl, %esi       ; This ignores all but the 5 lowest bits of %cl/%ecx
    call    printf

    movl    -4(%rbp), %ecx
    movl    $-1, %esi
    shrl    %cl, %esi
    call    printf