【发布时间】：2015-09-10 18:46:33
【问题描述】：
这是一项家庭作业。我有一个 C 程序，它调用一个用 NASM 实现的函数calc(int, float*, float*, float*, float*)。我想对从 C 传递的数据进行浮点除法，但首先我想检查我是否正确访问了数据。
这是C程序的节选：
printf("read.c: F data1[0]=%f\n", data1[0]);
printf("read.c: X data1[0]=%X\n", *(int*)(&data1[0]));
calc(nlines, data1, data2, result1, result2);
为了测试，我想从汇编代码中打印出完全相同的内容，但无论我尝试什么，它都不会给我正确的结果。准确地说，输出%X 格式会得到相同的结果，但%f 格式会给出一些令人难以置信的巨大数字。
global calc
extern printf
; -----------------------------------------------------------------------
; extern void calc(int nlines, float* data1, float* data2,
; float* result1, float* result2)
; -----------------------------------------------------------------------
calc:
section .data
.strf db "calc.asm: F data1[0]=%f", 10, 0
.strx db "calc.asm: X data1[0]=%X", 10, 0
section .text
enter 0, 0
; Move the value of float* data1 into ecx.
mov ecx, [esp + 12]
; Move the contents of data1[0] into esi.
mov esi, [ecx]
push esi
push .strf
call printf
add esp, 8
push esi
push .strx
call printf
add esp, 8
leave
ret
输出
read.c: F data1[0]=20.961977
read.c: X data1[0]=41A7B221
calc.asm: F data1[0]=-8796958457989122902187458235483374032941932827208012972482327255932202912296419757153331437662235555722313731094096197990916443553479942683040096290755684437514827018615169352974748429901549205109479495668937369584705401541113350145698235773041651907978442730240007381959397006695721667307435228446926569472.000000
calc.asm: X data1[0]=41A7B221
我也查看了fld，但我不知道如何将加载的值推送到堆栈上。这没有用：
; Move float* data1 into ecx
mov ecx, [esp + 12]
; Load the floating point number into esi.
fld dword [ecx]
fst esi
怎么做才对？
我已将 read.c 简化为此代码
#include <stdio.h>
#include <stdlib.h>
#define MAXLINES 1024
extern void calc(int, float*, float*, float*, float*);
int main(int argc, char** argv)
{
int nlines;
float* data1 = malloc(sizeof(float)*MAXLINES);
float*data2, *results1, *results2;
printf("read.c: F data1[0]=%f\n", data1[0]);
printf("read.c: X data1[0]=%X\n", *(int*)(&data1[0]));
calc(nlines, data1, data2, results1, results2);
return 0;
}
这是汇编程序的输出：
.file "test.c"
.section .rodata
.LC0:
.string "read.c: F data1[0]=%f\n"
.LC1:
.string "read.c: X data1[0]=%X\n"
.text
.globl main
.type main, @function
main:
.LFB2:
.cfi_startproc
pushl %ebp
.cfi_def_cfa_offset 8
.cfi_offset 5, -8
movl %esp, %ebp
.cfi_def_cfa_register 5
andl $-16, %esp
subl $64, %esp
movl $4096, (%esp)
call malloc
movl %eax, 44(%esp)
movl 44(%esp), %eax
flds (%eax)
fstpl 4(%esp)
movl $.LC0, (%esp)
call printf
movl 44(%esp), %eax
movl (%eax), %eax
movl %eax, 4(%esp)
movl $.LC1, (%esp)
call printf
movl 60(%esp), %eax
movl %eax, 16(%esp)
movl 56(%esp), %eax
movl %eax, 12(%esp)
movl 52(%esp), %eax
movl %eax, 8(%esp)
movl 44(%esp), %eax
movl %eax, 4(%esp)
movl 48(%esp), %eax
movl %eax, (%esp)
call calc
movl $0, %eax
leave
.cfi_restore 5
.cfi_def_cfa 4, 4
ret
.cfi_endproc
.LFE2:
.size main, .-main
.ident "GCC: (Ubuntu 4.8.4-2ubuntu1~14.04) 4.8.4"
.section .note.GNU-stack,"",@progbits
.LC1:
.string "read.c: F data1[0]=%f\n"
.LC2:
.string "read.c: X data1[0]=%X\n"
.text
.globl main
.type main, @function
main:
.LFB4:
.cfi_startproc
pushl %ebp
.cfi_def_cfa_offset 8
.cfi_offset 5, -8
movl %esp, %ebp
.cfi_def_cfa_register 5
andl $-16, %esp
subl $64, %esp
movl 44(%esp), %eax
flds (%eax)
fstpl 4(%esp)
movl $.LC1, (%esp)
call printf
movl 44(%esp), %eax
movl (%eax), %eax
movl %eax, 4(%esp)
movl $.LC2, (%esp)
call printf
movl 60(%esp), %eax
movl %eax, 16(%esp)
movl 56(%esp), %eax
movl %eax, 12(%esp)
movl 52(%esp), %eax
movl %eax, 8(%esp)
movl 44(%esp), %eax
movl %eax, 4(%esp)
movl 48(%esp), %eax
movl %eax, (%esp)
call calc
movl $0, %eax
leave
.cfi_restore 5
.cfi_def_cfa 4, 4
ret
.cfi_endproc
.LFE4:
.size main, .-main
.ident "GCC: (Ubuntu 4.8.4-2ubuntu1~14.04) 4.8.4"
.section .note.GNU-stack,"",@progbits
【问题讨论】：
-
IIRC，
%f期望双倍。因此，在将浮点数推入堆栈时，您需要将其转换为双精度数。像这样：fld dword [ecx]/sub esp,8/fstp qword [esp]。不要忘记相应地修改call printf之后的add。 -
你能发布一个 C 函数的反汇编吗？
-
@Michael 谢谢！但是，为什么当我为
%f传递浮点数而不是双精度时，C 中的 printf() 调用会起作用？ -
因为 float 参数被提升为双精度。
-
@AlexD 我已经更新了问题并添加了汇编器输出