Determining calling line without macro

Question

Is it possible to determine the line number that calls a function without the aid of a macro?

Consider this code:

#include <iostream>

#define PrintLineWithMacro() \
  std::cout << "Line: " << __LINE__ << std::endl;   // Line 4

void PrintLine()
{
  std::cout << "Line: " << __LINE__ << std::endl;   // Line 8
}

int main(int argc, char **argv)
{
  PrintLine();           // Line 13
  PrintLineWithMacro();  // Line 14
  return 0;
}

which outputs the following:

Line: 8
Line: 14

I understand why each prints what they do. I am more interested if it's possible to mimic the macro function without using a macro.

The only way to get the line is with `__LINE__`; you have to use the preprocessor somewhere. (In other words, the preprocessor deals with raw *code*, and line numbers are part of raw *code*.) — GManNickG, Oct 27 '10 at 01:20

score 6 · Accepted Answer · answered Oct 27 '10 at 01:19

6

I would do the following:

#define PrintLine() PrintLine_(__LINE__)

void PrintLine_(int line) {
    std::cout << "Line: " << line << std::endl;
}

I know that this doesn't completely remove the preprocessor, but it does move most of the logic into an actual function.

answered Oct 27 '10 at 01:19

Bill Lynch

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3

This is a decent workaround, because you *have* to use a macro. However, you should explicitly qualify the function call in the macro (::PrintLine_, or whichever namespace it's in) and use parentheses. – Oct 27 '10 at 01:28

score 2 · Answer 2 · answered Nov 16 '22 at 14:17

C++ has caught up with this question in the c++20 standard.

In c++20 you may now do:

#include <iostream>
#include <source_location>

void PrintLine(std::source_location location = std::source_location::current())
{
    std::cout << "Line: " << location.line() << std::endl;   // Line 8
}

int main(int argc, char **argv)
{
    PrintLine();           // Line: 11
    PrintLine();           // Line: 12
    return 0;
}

Another question is whether the compiler of your interest has caught up with the standard. Note that the construct was present pre c++20 and you could try:

#include <experimental/source_location>

score 1 · Answer 3 · answered Oct 27 '10 at 01:22

Not portably. On any given platform, you could basically re-implement the details of a debugger - the information is effectively stored on your stack as the return address. You can get at that kind of thing with the backtrace() function on some platforms.

Determining calling line without macro

3 Answers3