所以我有一个简单的C程序循环传递给main的args然后返回:
#include <stdio.h>
int main(int argc, char *argv[])
{
int i;
for(i = 0; i < argc; ++i) {
fprintf(stdout, "%s\n", argv[i]);
}
return 0;
}
我想看看gcc如何以NASM格式写出程序集。我正在查看.asm文件中的输出,并注意到语法是TASM。下面是make文件和gcc的输出。我做错了还是gcc没有输出真正的NASM语法?
all: main
main: main.o
ld -o main main.o
main.o : main.c
gcc -S -masm=intel -o main.asm main.c
nasm -f elf -g -F stabs main.asm -l main.lst
和
.file "main.c"
.intel_syntax noprefix
.section .rodata
.LC0:
.string "%s\n"
.text
.globl main
.type main, @function
main:
push ebp
mov ebp, esp
and esp, -16
sub esp, 32
mov DWORD PTR [esp+28], 0
jmp .L2
.L3:
mov eax, DWORD PTR [esp+28]
sal eax, 2
add eax, DWORD PTR [ebp+12]
mov ecx, DWORD PTR [eax]
mov edx, OFFSET FLAT:.LC0
mov eax, DWORD PTR stdout
mov DWORD PTR [esp+8], ecx
mov DWORD PTR [esp+4], edx
mov DWORD PTR [esp], eax
call fprintf
add DWORD PTR [esp+28], 1
.L2:
mov eax, DWORD PTR [esp+28]
cmp eax, DWORD PTR [ebp+8]
jl .L3
mov eax, 0
leave
ret
.size main, .-main
.ident "GCC: (GNU) 4.5.1 20100924 (Red Hat 4.5.1-4)"
.section .note.GNU-stack,"",@progbits
命令行上的错误是:
[mehoggan@fedora sandbox-print_args]$ make
gcc -S -masm=intel -o main.asm main.c
nasm -f elf -g -F stabs main.asm -l main.lst
main.asm:1: error: attempt to define a local label before any non-local labels
main.asm:1: error: parser: instruction expected
main.asm:2: error: attempt to define a local label before any non-local labels
main.asm:2: error: parser: instruction expected
main.asm:3: error: attempt to define a local label before any non-local labels
main.asm:3: error: parser: instruction expected
main.asm:4: error: attempt to define a local label before any non-local labels
main.asm:5: error: attempt to define a local label before any non-local labels
main.asm:5: error: parser: instruction expected
main.asm:6: error: attempt to define a local label before any non-local labels
main.asm:7: error: attempt to define a local label before any non-local labels
main.asm:7: error: parser: instruction expected
main.asm:8: error: attempt to define a local label before any non-local labels
main.asm:8: error: parser: instruction expected
main.asm:14: error: comma, colon or end of line expected
main.asm:17: error: comma, colon or end of line expected
main.asm:19: error: comma, colon or end of line expected
main.asm:20: error: comma, colon or end of line expected
main.asm:21: error: comma, colon or end of line expected
main.asm:22: error: comma, colon or end of line expected
main.asm:23: error: comma, colon or end of line expected
main.asm:24: error: comma, colon or end of line expected
main.asm:25: error: comma, colon or end of line expected
main.asm:27: error: comma, colon or end of line expected
main.asm:29: error: comma, colon or end of line expected
main.asm:30: error: comma, colon or end of line expected
main.asm:35: error: parser: instruction expected
main.asm:36: error: parser: instruction expected
main.asm:37: error: parser: instruction expected
make: *** [main.o] Error 1
是什么让我相信这是TASM语法是在此链接上发布的信息: http://rs1.szif.hu/~tomcat/win32/intro.txt
TASM编码员通常在NASM中存在词汇困难,因为它 缺少在TASM中广泛使用的“ptr”关键字。
TASM使用此:
mov al,byte ptr [ds:si]或mov ax,word ptr [ds:si]或mov eax, dword ptr [ds:si]
对于NASM这只是转换为:
mov al,byte [ds:si]或mov ax,word [ds:si]或mov eax,dword [DS:SI]
NASM允许在很多地方使用这些大小的关键字,从而为您提供了一个 以统一的方式对生成的操作码进行大量控制 示例这些都是有效的:
push dword 123 jmp [ds:word 1234];这些都指定了大小 偏移量jmp [ds:dword 1234];对于棘手的代码 接口32位和 ; 16位段
它会变得很毛茸茸,但重要的是要记住你 你可以随时拥有所需的所有控制权。
答案 0 :(得分:7)
Intel语法是指Intel语法,而不是NASM语法。 MASM和TASM语法基于英特尔语法,NASM语法从英特尔语法中获得灵感,但它有所不同。
gcc输出实际上是使用英特尔语法进行单独指令的气体语法,(汇编指令,标签等使用气体特定语法)