__VA_ARGS__, is this code snippet portable?

Question

Is this code snippet portable?

definition:

#define log(...)  std::cout << __FILE__ << "[" << __LINE__  << "]:" \
                            << string_format(__VA_ARGS__) \
                            << std::endl << std::flush

usage:

log("i = %d", i);

and string_format is sprintf that outputs a std::string taken from https://stackoverflow.com/a/8098080/624074 It works with my gcc 4.6.3 here but I don't want to have compilation issues later with other compilers.

You are right not determine portability by empirical test. But why print log messages to `cout` and not `clog`? And `endl` already performs a flush. — Potatoswatter, Jan 09 '14 at 02:36
what is clog? Is endl flush part of standard or compiler dependent? — Mohammad Moghimi, Jan 09 '14 at 02:45

score 2 · Accepted Answer · edited Jun 20 '20 at 09:12

2

Yes, the macro __VA_ARGS__ is part of standard C++:

C++11 §16.3.1 Argument substitution

An identifier __VA_ARGS__that occurs in the replacement list shall be treated as if it were a parameter, and the variable arguments shall form the preprocessing tokens used to replace it.

edited Jun 20 '20 at 09:12

Community

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answered Jan 09 '14 at 02:23

Yu Hao

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score 2 · Answer 2 · answered Jan 09 '14 at 02:24

2

It's pretty portable, for the most part. It's part of the C++11 language standard (as well as C99), so most modern compilers support it. You might run into problems with older compilers that only support C++03.

answered Jan 09 '14 at 02:24

Adam Rosenfield

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__VA_ARGS__, is this code snippet portable?

2 Answers2

C++11 §16.3.1 Argument substitution