首先，您的问题的另一个答案是错误的：在Linux上，您可以从链接器获取文件和行号：

First, the other answer to your question is wrong: on Linux you do get file and line number from the linker:

$ cat foo.cc
extern int myFunction(void);

int main()
{
  return myFunction();
}
$ g++ -g foo.cc
/tmp/cc3twlhL.o: In function `main':
/tmp/foo.cc:5: undefined reference to `myFunction()'
collect2: ld returned 1 exit status

c $ c> gcc（Ubuntu / Linaro 4.6.3-1ubuntu5）4.6.3 和链接器 GNU ld（GNU Binutils for Ubuntu）2.22 但是对于更旧版本的GCC和ld也是如此。

Above output is from gcc (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3 and linker GNU ld (GNU Binutils for Ubuntu) 2.22, but this has been true for much older versions of GCC and ld as well.

无法获取文件/行的原因必须是

The reason you are not getting the file/line must be that

• 您未使用 -g 标志或


• 您拥有真的旧 ld 或


• 您已配置 ld 不支持调试（我不确定这是否可能）。

• you didn't use -g flag, or
• you have a really old ld, or
• you have configured your ld without support for debugging (I am not sure this is even possible).

ld 拒绝告诉您文件和行，不是所有都丢失。您可以将源代码编译成对象，然后使用 objdump -rdS foo.o 获得相同的信息：

However, even if your ld is refusing to tell you the file and line, not all is lost. You can compile your source into object, then use objdump -rdS foo.o to obtain the same info:

g++ -g -c foo.cc
objdump -rdS foo.o

Disassembly of section .text:

0000000000000000 <main>:
extern int myFunction(void);

int main()
{
   0:   55                      push   %rbp
   1:   48 89 e5                mov    %rsp,%rbp
  return myFunction();
   4:   e8 00 00 00 00          callq  9 <main+0x9>
            5: R_X86_64_PC32    _Z10myFunctionv-0x4
}
   9:   5d                      pop    %rbp
   a:   c3                      retq

在上面的输出中，您可以清楚地看到哪条源代码行引用了 _Z10myFunctionv （这是 C ++ myFunction（void）的错误名称将在对象文件中发出。

In above output, you can clearly see which source line caused reference to _Z10myFunctionv (which is the C++ mangled name for myFunction(void)) to be emitted in the object file.