32 位机器如何处理大于 2^32 的数字? [英] How are numbers greater than 2^32 handled by a 32 bit machine?

问题描述

我试图了解涉及大于 2³² 的数字的计算是如何在 32 位机器上发生的.

I am trying to understand how calculations involving numbers greater than 2³² happen on a 32 bit machine.

C 代码

$ cat size.c
#include<stdio.h>
#include<math.h>

int main() {

    printf ("max unsigned long long = %llu
",
    (unsigned long long)(pow(2, 64) - 1));
}
$

gcc 输出

$ gcc size.c -o size
$ ./size
max unsigned long long = 18446744073709551615
$

对应的汇编代码

$ gcc -S size.c -O3
$ cat size.s
    .file   "size.c"
    .section    .rodata.str1.4,"aMS",@progbits,1
    .align 4
.LC0:
    .string "max unsigned long long = %llu
"
    .text
    .p2align 4,,15
.globl main
    .type   main, @function
main:
    pushl   %ebp
    movl    %esp, %ebp
    andl    $-16, %esp
    subl    $16, %esp
    movl    $-1, 8(%esp)   #1
    movl    $-1, 12(%esp)  #2
    movl    $.LC0, 4(%esp) #3
    movl    $1, (%esp)     #4
    call    __printf_chk
    leave
    ret
    .size   main, .-main
    .ident  "GCC: (Ubuntu 4.4.3-4ubuntu5) 4.4.3"
    .section    .note.GNU-stack,"",@progbits
$

第 1 - 4 行究竟发生了什么?

这是汇编级别的某种字符串连接吗?

Is this some kind of string concatenation at the assembly level?

推荐答案

__printf_chk 是 printf 的包装器，它检查堆栈溢出，并采用额外的第一个参数，一个标志(例如，参见 此处.)

__printf_chk is a wrapper around printf which checks for stack overflow, and takes an additional first parameter, a flag (e.g. see here.)

pow(2, 64) - 1 已优化为 0xffffffffffffffff，因为参数是常量.

pow(2, 64) - 1 has been optimised to 0xffffffffffffffff as the arguments are constants.

按照通常的调用约定，__printf_chk() (int flag) 的第一个参数是堆栈上的 32 位值(位于 %esp 在 call 指令时).下一个参数 const char * format 是一个 32 位指针(堆栈中的下一个 32 位字，即在 %esp+4 处).并且正在打印的 64 位数量占据接下来的两个 32 位字(在 %esp+8 和 %esp+12 处):

As per the usual calling conventions, the first argument to __printf_chk() (int flag) is a 32-bit value on the stack (at %esp at the time of the call instruction). The next argument, const char * format, is a 32-bit pointer (the next 32-bit word on the stack, i.e. at %esp+4). And the 64-bit quantity that is being printed occupies the next two 32-bit words (at %esp+8 and %esp+12):

pushl   %ebp                 ; prologue
movl    %esp, %ebp           ; prologue
andl    $-16, %esp           ; align stack pointer
subl    $16, %esp            ; reserve bytes for stack frame
movl    $-1, 8(%esp)   #1    ; store low half of 64-bit argument (a constant) to stack
movl    $-1, 12(%esp)  #2    ; store high half of 64-bit argument (a constant) to stack
movl    $.LC0, 4(%esp) #3    ; store address of format string to stack
movl    $1, (%esp)     #4    ; store "flag" argument to __printf_chk to stack
call    __printf_chk         ; call routine
leave                        ; epilogue
ret                          ; epilogue

编译器有效地重写了这个:

The compiler has effectively rewritten this:

printf("max unsigned long long = %llu
", (unsigned long long)(pow(2, 64) - 1));

...进入这个:

__printf_chk(1, "max unsigned long long = %llu
", 0xffffffffffffffffULL);

...而且，在运行时，调用的堆栈布局如下所示(将堆栈显示为 32 位字，地址从图表底部向上增加):

...and, at runtime, the stack layout for the call looks like this (showing the stack as 32-bit words, with addresses increasing from the bottom of the diagram upwards):

        :                 :
        :     Stack       :
        :                 :
        +-----------------+
%esp+12 |      0xffffffff |  
        +-----------------+  } <-------------------------------------.
%esp+8  |      0xffffffff | /                                        |
        +-----------------+                                          |
%esp+4  |address of string| <---------------.                        |
        +-----------------+                 |                        |
%esp    |               1 | <--.            |                        |
        +-----------------+    |            |                        |
                  __printf_chk(1, "max unsigned long long = %llu
", |
                                                    0xffffffffffffffffULL);

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