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Editorial Team
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Asked: 2026-06-15T01:55:25+00:00

The following code compiles fine on Windows with Visual Studio: class_handle(base *ptr) : ptr_m(ptr),

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The following code compiles fine on Windows with Visual Studio:

class_handle(base *ptr) : ptr_m(ptr), name_m(typeid(base).raw_name()) { signature_m = CLASS_HANDLE_SIGNATURE; }

If I try to compile the same code on Linux I get:

error: ‘const class std::type_info’ has no member named ‘raw_name’

as far as I understand, raw_name is a Microsoft specific implementation. How do I have to change my code so it compiles both on Windows and Linux systems?

EDIT1 I prefer to not modify the original code, I just need a workaround to compile with gcc. Is that possible?

EDIT2 will #define raw_name name do the trick?

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1 Answer

  1. Editorial Team

    Write these:

    // for variables:
template<typename T>
char const* GetRawName( T unused ) { ... }
// for types:
template<typename T>
char const* GetRawName() { ... }

    with different implementation on Windows and not-on-Windows using an #ifdef block on a token you know to be defined in the microsoft compiler, but not in your other compiler. This isolates the preprocessing differences between MS and non-MS compiled versions to an isolated file.

    This does require a minimal amount of change to the original code, but does so in a way that will still compile on the microsoft compiler.

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