CODEWITHSUNDEEP
c++macros
Petr
Petr

Reputation: 14455

What is a best way to check if compiler supports __FUNC__ etc

There are some special macros supported by most compilers, such as __FUNC__ or __PRETTY_FUNCTION__ but I am wondering how can I verify if these are available during compile time, there is this Boost macro from boost/current_function.hpp:

#if defined(__GNUC__) || (defined(__MWERKS__) && (__MWERKS__ >= 0x3000)) || (defined(__ICC) && (__ICC >= 600))

# define BOOST_CURRENT_FUNCTION __PRETTY_FUNCTION__

#elif defined(__DMC__) && (__DMC__ >= 0x810)

# define BOOST_CURRENT_FUNCTION __PRETTY_FUNCTION__

#elif defined(__FUNCSIG__)

# define BOOST_CURRENT_FUNCTION __FUNCSIG__

#elif (defined(__INTEL_COMPILER) && (__INTEL_COMPILER >= 600)) || (defined(__IBMCPP__) && (__IBMCPP__ >= 500))

# define BOOST_CURRENT_FUNCTION __FUNCTION__

#elif defined(__BORLANDC__) && (__BORLANDC__ >= 0x550)

# define BOOST_CURRENT_FUNCTION __FUNC__

#elif defined(__STDC_VERSION__) && (__STDC_VERSION__ >= 199901)

# define BOOST_CURRENT_FUNCTION __func__

#else

# define BOOST_CURRENT_FUNCTION "(unknown)"

#endif

However, does it really need to be so complex? Isn't it possible to #ifdef __PRETTY_FUNCTION__ instead so that it would use this macro even on compilers that are not known by boost or the person who wrote the macro? For example:

#if defined(__PRETTY_FUNCTION__)

# define BOOST_CURRENT_FUNCTION __PRETTY_FUNCTION__

#elif defined(__FUNCSIG__)

# define BOOST_CURRENT_FUNCTION __FUNCSIG__

#elif (defined(__FUNCTION__)

# define BOOST_CURRENT_FUNCTION __FUNCTION__

#elif defined(__FUNC__)

# define BOOST_CURRENT_FUNCTION __FUNC__

#elif defined(__func__)

# define BOOST_CURRENT_FUNCTION __func__

#else

# define BOOST_CURRENT_FUNCTION "(unknown)"

#endif

What is a difference? Would it work on all compilers?

Upvotes: 0

Views: 843

Answers (1)

Allison Lock
Allison Lock

Reputation: 2383

See https://stackoverflow.com/a/4384825/583195.

It quotes:

The identifier __func__ is implicitly declared by the translator as if, immediately following the opening brace of each function definition, the declaration

static const char __func__[] = "function-name";

appeared, where function-name is the name of the lexically-enclosing function. This name is the unadorned name of the function.

This indicates that __func__ and many of the other "magic macros" are not macros at all, and therefore not picked up by #ifdef

Upvotes: 2

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