我正在尝试编写一个简单的程序，在不使用linux系统调用或标准C库函数的情况下打印出一个C字符串。这仅用于学习目的，我绝不会在制作中这样做（除非我真的很擅长它=））。
首先我的系统信息：
[mehoggan@fedora sandbox-print_chars]$ uname -a
Linux fedora.laptop 2.6.35.14-106.fc14.i686.PAE #1 SMP Wed Nov 23 13:39:51 UTC 2011 i686 i686 i386 GNU/Linux
[mehoggan@fedora sandbox-print_chars]$ gcc --version
gcc (GCC) 4.5.1 20100924 (Red Hat 4.5.1-4)
Copyright (C) 2010 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
接下来是代码：
#include <unistd.h>
#include <sys/syscall.h>
void main()
{
char *str = "Hello World";
while(*(str) != '\0') {
//printf("%c", *(str++));
//syscall(__NR_write, 1, *(str++), 1);
__asm__( "movl %0, %%ecx" :"=c" (str));
__asm__( "movl $0X4, %eax" );
__asm__( "movl $0X1, %ebx" );
__asm__( "movl $0X1, %edx" );
__asm__( "int $0X80" );
str++;
}
return;
}
使用以下makefile编译：
all: sandbox_c
sandbox_c: sandbox.c
gcc -Wall -o sandbox_c ./sandbox.c
gcc -S -Wall -o sandbox_c.asm ./sandbox.c
ADDITION 通过gdb运行可执行文件我可以看到ecx没有指向正确的地址：
[mehoggan@fedora sandbox-print_chars]$ gdb ./sandbox_c
GNU gdb (GDB) Fedora (7.2-52.fc14)
Copyright (C) 2010 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law. Type "show copying"
and "show warranty" for details.
This GDB was configured as "i686-redhat-linux-gnu".
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>...
Reading symbols from /home/mehoggan/Code/Assembly/sandbox/sandbox-print_chars/sandbox_c...done.
(gdb) break sandbox.c:8
Breakpoint 1 at 0x80483a2: file ./sandbox.c, line 8.
(gdb) run
Starting program: /home/mehoggan/Code/Assembly/sandbox/sandbox-print_chars/sandbox_c
Breakpoint 1, main () at ./sandbox.c:8
8 while(*(str) != '\0') {
Missing separate debuginfos, use: debuginfo-install glibc-2.13-2.i686
(gdb) step
11 __asm__( "movl %0, %%ecx" :"=c" (str));
(gdb) info registers
eax 0x48 72
ecx 0x34092fad 873017261
edx 0x1 1
ebx 0x567ff4 5668852
esp 0xbffff2a4 0xbffff2a4
ebp 0xbffff2b8 0xbffff2b8
esi 0x0 0
edi 0x0 0
eip 0x80483a4 0x80483a4 <main+16>
eflags 0x200206 [ PF IF ID ]
cs 0x73 115
ss 0x7b 123
ds 0x7b 123
es 0x7b 123
fs 0x0 0
gs 0x33 51
(gdb) step
12 __asm__( "movl $0X4, %eax" );
(gdb) info registers
eax 0x48 72
ecx 0x34092fad 873017261
edx 0x1 1
ebx 0x34092fad 873017261
esp 0xbffff2a4 0xbffff2a4
ebp 0xbffff2b8 0xbffff2b8
esi 0x0 0
edi 0x0 0
eip 0x80483ab 0x80483ab <main+23>
eflags 0x200206 [ PF IF ID ]
cs 0x73 115
ss 0x7b 123
ds 0x7b 123
es 0x7b 123
fs 0x0 0
gs 0x33 51
(gdb) step
13 __asm__( "movl $0X1, %ebx" );
(gdb) info registers
eax 0x4 4
ecx 0x34092fad 873017261
edx 0x1 1
ebx 0x34092fad 873017261
esp 0xbffff2a4 0xbffff2a4
ebp 0xbffff2b8 0xbffff2b8
esi 0x0 0
edi 0x0 0
eip 0x80483b0 0x80483b0 <main+28>
eflags 0x200206 [ PF IF ID ]
cs 0x73 115
ss 0x7b 123
ds 0x7b 123
es 0x7b 123
fs 0x0 0
gs 0x33 51
(gdb) step
14 __asm__( "movl $0X1, %edx" );
(gdb) info registers
eax 0x4 4
ecx 0x34092fad 873017261
edx 0x1 1
ebx 0x1 1
esp 0xbffff2a4 0xbffff2a4
ebp 0xbffff2b8 0xbffff2b8
esi 0x0 0
edi 0x0 0
eip 0x80483b5 0x80483b5 <main+33>
eflags 0x200206 [ PF IF ID ]
cs 0x73 115
ss 0x7b 123
ds 0x7b 123
es 0x7b 123
fs 0x0 0
gs 0x33 51
(gdb) step
15 __asm__( "int $0X80" );
(gdb) info registers
eax 0x4 4
ecx 0x34092fad 873017261
edx 0x1 1
ebx 0x1 1
esp 0xbffff2a4 0xbffff2a4
ebp 0xbffff2b8 0xbffff2b8
esi 0x0 0
edi 0x0 0
eip 0x80483ba 0x80483ba <main+38>
eflags 0x200206 [ PF IF ID ]
cs 0x73 115
ss 0x7b 123
ds 0x7b 123
es 0x7b 123
fs 0x0 0
gs 0x33 51
(gdb) step
16 str++;
(gdb) info registers
eax 0xfffffff2 -14
ecx 0x34092fad 873017261
edx 0x1 1
ebx 0x1 1
esp 0xbffff2a4 0xbffff2a4
ebp 0xbffff2b8 0xbffff2b8
esi 0x0 0
edi 0x0 0
eip 0x80483bc 0x80483bc <main+40>
eflags 0x200206 [ PF IF ID ]
cs 0x73 115
ss 0x7b 123
ds 0x7b 123
es 0x7b 123
fs 0x0 0
gs 0x33 51
答案 0 :(得分：2)
试试这个：
__asm__ volatile ( "movl $0X4, %eax
movl $0X1, %ebx
movl $0X1, %edx
int $0X80"
: /* outputs: */ /* none */
: /* inputs: */ "c" (str)
: /* clobbers: */ "eax", "ebx", "edx");
我没有测试过，但语法看起来不错。如果系统调用覆盖其他任何内容，您可能需要添加更多“clobbers” - 请查看文档。
打破它：
%0移至%ecx，因为"c"约束已经执行了此操作。str是一个输入，但您已将其作为输出。volatile告诉编译器不要删除它 - 它没有输出，所以编译器可能认为它可以。答案 1 :(得分：0)
无论它值多少，调用“int 0x80” 进行系统调用。
您只是没有使用“printf”，或者使用Linux系统调用的标准C库包装器。
这并没有错：:)
无论如何，这里有一个完整的例子，说明你正在追求的是什么：
http://asm.sourceforge.net/intro/hello.html
'希望有帮助......并且玩得开心！