我只是想编译一些我从书中输入的简单示例代码,GCC给出了上述错误 . 这是我的代码:
$ cat -n test.cpp
1 #define READ_COMMAND 3
2
3 #define MSG_LENGTH 128
4
5 #include <stdlib.h>
6 #include <stdio.h>
7
8 int main(int argc, char *arg[])
9 {
10 int syslog_command = READ_COMMAND;
11 int bytes_to_read = MSG_LENGTH;
12 int retval;
13 char buffer[MSG_LENGTH];
14
15 asm volatile(
16 "movl %1, %%ebx\n\t"
17 "movl %2, %%ecx\n\t"
18 "movl %3, %%edx\n\t"
19 "movl $103, %%eax\n\t"
20 "int $128\n\t"
21 "movl %%eax, %0"
22 :"=r" (retval)
23 :"m"(syslog_command),"r"(buffer),"m"(bytes_to_read)
24 :"%eax","%ebx","%ecx","%edx");
25 if (retval > 0) printf("%s\n", buffer);
26
27 }
28
29
该代码应该调用 syslog()
系统调用以从内核 printk()
环形缓冲区读取最后128个字节 . 以下是有关我的操作系统和系统配置的一些信息:
uname -a
:
Linux 3.2.0-26-generic#41-Ubuntu SMP Thu Jun 14 17:49:24 UTC 2012 x86_64 x86_64 x86_64 GNU / Linux
gcc -v
:
Using built-in specs.
COLLECT_GCC=gcc
COLLECT_LTO_WRAPPER=/usr/lib/gcc/x86_64-linux-gnu/4.6/lto-wrapper
Target: x86_64-linux-gnu
配置为:../ src / configure -v --with-pkgversion ='Ubuntu / Linaro 4.6.3-1ubuntu5'--with-bugurl = file:///usr/share/doc/gcc-4.6/README .Bugs --enable-languages = c,c,fortran,objc,obj-c --prefix = / usr --program-suffix = -4.6 --enable-shared --enable-linker-build-id --with -system-zlib --libexecdir = / usr / lib --without-included-gettext --enable-threads = posix --with-gxx-include-dir = / usr / include / c /4.6 --libdir = / usr / lib --enable-nls --with-sysroot = / --enable-clocale = gnu --enable-libstdcxx-debug --enable-libstdcxx-time = yes --enable-gnu-unique-object --enable-插件--enable-objc-gc --disable-werror --with-arch-32 = i686 --with-tune = generic --enable-checking = release --build = x86_64-linux-gnu --host = x86_64 -linux-gnu --target = x86_64-linux-gnu
Thread model: posix
gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5)
下面是完整的错误:
$ gcc test.cpp
test.cpp: Assembler messages:
test.cpp:25: Error: unsupported for `mov'
1 回答
您正在尝试在64位计算机上编译32位汇编代码 . 您列出的内联汇编编译为:
如您所见,您试图在32位寄存器中存储64位寄存器,这是非法的 . 更重要的是,这也不是64位ABI系统调用协议 .
尝试使用
-m32
进行编译以强制使用32位ABI .