问题描述
我试图了解在32位计算机上如何进行大于2 的数字计算。
I am trying to understand how calculations involving numbers greater than 2 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\n",
(unsigned long long)(pow(2, 64) - 1));
}
$
gcc输出
$ gcc size.c -o size
$ ./size
max unsigned long long = 18446744073709551615
$
相应的汇编代码
$ 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\n"
.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行到底发生了什么?
这是某种在汇编级别的字符串连接吗?
Is this some kind of string concatenation at the assembly level?
推荐答案
__ printf_chk
是 printf
的包装,它检查堆栈是否溢出,并接受一个附加的第一个参数a标志(例如参见。)
__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
作为参数是常量。
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的第一个参数()
( int标志
)是堆栈上的32位值(位于%esp
调用
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%esp + 4
)。接下来要打印的64位数字占据了接下来的两个32位字(分别为%esp + 8
和%esp + 12
):
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
编译器实际上已将其重写为:
The compiler has effectively rewritten this:
printf("max unsigned long long = %llu\n", (unsigned long long)(pow(2, 64) - 1));
...到此:
__printf_chk(1, "max unsigned long long = %llu\n", 0xffffffffffffffffULL);
...在运行时,调用的堆栈布局如下所示(显示堆栈作为32位字,地址从图的底部开始逐渐增加):
...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\n", |
0xffffffffffffffffULL);
这篇关于32位机器如何处理大于2 ^ 32的数字?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!