Question

我只是想编译一些我从书中输入的简单示例代码，GCC给了我上面的错误。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

代码应该调用syslog()系统调用来读取内核printk()环形缓冲区中的最后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-plugin --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'

Answer 1

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：

movl -24(%rbp), %ebx
movl %rsi, %ecx       <--- error here
movl -28(%rbp), %edx
movl $103, %eax
int $128
movl %eax, %r12d

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。

GCC错误消息“错误：不支持'mov'”是什么意思？

1 个答案: