移位运算溢出：右操作数须小于左操作数的位数

Posted 2020-10-31 枝桠

#include <stdio.h>

int main(){
        int x = 1 << 32;
        int y = 32;
        int z = 1 << y;
        printf("x:%d, z: %d\n", x, z);
}

以上运行结果： x:0, z: 1 

汇编代码如下：

 1         .file   "tmp.c"
 2         .section        .rodata
 3 .LC0:
 4         .string "x:%d, z: %d\n"
 5         .text
 6         .globl  main
 7         .type   main, @function
 8 main:
 9 .LFB0:
10         .cfi_startproc
11         pushq   %rbp
12         .cfi_def_cfa_offset 16
13         .cfi_offset 6, -16
14         movq    %rsp, %rbp
15         .cfi_def_cfa_register 6
16         subq    $16, %rsp
17         movl    $0, -4(%rbp)            ; ?
18         movl    $32, -8(%rbp)           ; y = 32;
19         movl    -8(%rbp), %eax          ; $32 => %eax
20         movl    $1, %edx                        ; $1 => %edx
21         movl    %eax, %ecx                      ; $32 => %ecx
22         sall    %cl, %edx                       ; 1 << 32
23         movl    %edx, %eax                      ; (1 << 32) => %eax
24         movl    %eax, -12(%rbp)         ; z
25         movl    -12(%rbp), %edx         ; z => %edx
26         movl    -4(%rbp), %eax          ; $0 => %eax
27         movl    %eax, %esi
28         movl    $.LC0, %edi
29         movl    $0, %eax
30         call    printf
31         leave
32         .cfi_def_cfa 7, 8
33         ret
34         .cfi_endproc
35 .LFE0:
36         .size   main, .-main
37         .ident  "GCC: (GNU) 4.8.5 20150623 (Red Hat 4.8.5-16)"
38         .section        .note.GNU-stack,"",@progbits

 

看汇编代码，  1 << 32 和 1 << y 明显是不一样的机制实现的， 虽然我也看不懂~~~

 

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这个问题也是在Stack Overflow上看到的，原文地址https://stackoverflow.com/questions/3871650/gcc-left-shift-overflow