如何反编译这个x87装配计算？

Question

我正在反转的程序在浮点数和8字节整数之间进行简单的乘法运算:

section .data

va: dt 1.4426950408889634074
vb: dd 0x42424242
    dd 0x41414141

section .text
global main

main:
    fld tword[va]
    fmul qword[vb]
    ret

gdb下的结果:

Breakpoint 1, 0x08048360 in main ()
(gdb) x/i $eip
0x8048360 <main>:       fld    TBYTE PTR ds:0x804953c
0x8048366 <main+6>:     fmul   QWORD PTR ds:0x8049546
0x804836c <main+12>:    ret
(gdb) x/gx 0x8049546
0x8049546 <vb>: 0x4141414142424242
(gdb) si
0x08048366 in main ()
0x0804836c in main ()
(gdb) info float
=>R7: Valid   0x4014c726039c95268dc4 +3262848.902912714389

我正在尝试在C(相同的32位环境)中重新创建此程序:

#include <stdio.h>

int main() {

    unsigned long long vb = 0x4141414142424242LL;
    float r, va = 1.4426950408889634074F;

    r = va * vb;
    printf("%f\n", r);
}

...但我得到了非常不同的结果:

$ ./test
6783712964982603776.000000

我在C程序中做错了什么？

Answer 1

在asm代码中,你实际上将两个doubles与fmul指令相乘,而不是a float和a int.在C中做类似的事情:

#include <stdio.h>
#include <stdint.h>
#include <string.h>

int main()
{
    uint64_t vbi = 0x4141414142424242ULL; // hex representation of double
    double r, vb, va = 1.4426950408889634074;

    memcpy(&vb, &vbi, sizeof(vb));        // copy hex to double
    r = va * vb;
    printf("va = %f, vb = %f, r = %f\n", va, vb, r);
    return 0;
}

结果= va = 1.442695, vb = 2261634.517647, r = 3262848.902913.

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