c++ - How to go from linker error to line of code in the sources? -

the linker produces kind of output

/var/tmp/ccitb4j2.o: in function `main': /var/tmp/ccitb4j2.o(.text+0x4): undefined reference `myfunction(void)' 

how can find out line of source code corresponding instruction @ .text+0x4 function invoked?

first, other answer question wrong: on linux do file , line number 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 `myfunction()' collect2: ld returned 1 exit status 

above output gcc (ubuntu/linaro 4.6.3-1ubuntu5) 4.6.3 , linker gnu ld (gnu binutils ubuntu) 2.22, has been true older versions of gcc , ld well.

the reason not getting file/line must that

• you didn't use -g flag, or
• you have really old ld, or
• you have configured ld without support debugging (i not sure possible).

however, if ld refusing tell file , line, not lost. can compile source object, use objdump -rds foo.o obtain 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 

in above output, can see source line caused reference _z10myfunctionv (which c++ mangled name myfunction(void)) emitted in object file.