At what point in the preprocessing phase does __LINE__ expand?

Question

I have a function in the header file:

template <int line>
inline void log() {}

And then I try this trick to make it easier to use:

#define LOG_LINE log<__LINE__>()

And then in the .cpp file I:

void main()
{
    LOG_LINE;
}

And it looks like it works the way I would like. I get the line from the .cpp file, not the line in which it is LOG_LINEdeclared in the .h file. But I don’t understand how it works. Does preprocessing do dual-processing C ++, leaving behind special macros such as __LINE__for the second pass? Is this a portable (standard) behavior? Should I expect this to work with all major C ++ compilers? So far I have only tried MSVC.

Answer 1

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Answer 2

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, LOG_LINE log<__LINE__>(), __LINE__ 3, log<3>() - .

Answer 3

This does not require double pass pretreatment. We are talking about the extension order of nested macros (well, in a sense, the expression expands in as many passes as necessary to expand all nested macros). Quote from the answer to this question :

All arguments that do not appear after the # operator or any of the ## sides expand completely when they are replaced, and not before a functionally similar macro expands.