How to get the file name and a line number of a function call?
c++
Solution
In general you can do this automatically by hiding your function behind a macro call which passes allong the `__FILE__` and `__LINE__` values
void _B(const char* file, int line) { ... }
#define B() _B(__FILE__, __LINE__)
This is by no means a foolproof solution though. It's possible for developers to call `_B` directly or for `_B` to be called from generated code, assembly, etc .... where there may be no meaningful file / line number
OP asked for an example with arguments
void _C(int p1, char p2, const char* file, int line) { ... }
#define C(p1, p2) _C(p1, p2, __FILE__, __LINE__)
Problem
I have two functions in different source files: a.cpp ``` void A() { B(); } ``` b.cpp ``` void B() { std::cout << "B() called from file: " << ??? << " line: " << ??? << std::endl; } ``` How can I get the file name and line number of the call?