is there a way i can print __FUNCTION__
as a wide character on linux?
the trick with the WIDEN doesn't work for me, the gcc compiler prints: error: ?L_FUNCTION_? was not declared in this scope
any help? Thanks
is there a way i can print __FUNCTION__
as a wide character on linux?
the trick with the WIDEN doesn't work for me, the gcc compiler prints: error: ?L_FUNCTION_? was not declared in this scope
any help? Thanks
7+ years later (although things seem to be the same) ...
Since you mentioned gcc, check [GNU.GCC]: Standard Predefined Macros (emphasis is mine):
C99 introduced __func__, and GCC has provided __FUNCTION__ for a long time. Both of these are strings containing the name of the current function (there are slight semantic differences; see the GCC manual). Neither of them is a macro; the preprocessor does not know the name of the current function.
Since __FUNCTION__ is not a macro (the preprocessor doesn't "know" anything about it), it will remain untouched during (outer) macro expansion, generating at the end the L__FUNCTION__
identifier, which is obviously invalid.
That's why the double macro approach works for __FILE__ (for example), but not for __FUNCTION__ (or __func__).
So, the short answer to your question is "NO" (at least not at preprocessor level). You will need to convert __FUNCTION__ "manually" (e.g. using one of the [man7]: MBSTOWCS(3) functions family).
Note: It works on VStudio, since according to [MS.Docs]: Predefined Macros (emphasis still mine):
- __FUNCTION__ Defined as a string literal that contains the undecorated name of the enclosing function. The macro is defined only within a function.
It can be done using macros, you just have to understand how macros expand. To get a wide char version of the macro you need to create 2 layers of macros like so:
#define WIDE2(x) L##x
#define WIDECHAR(x) WIDE2(x)
#define WIDE_FUNCTION WIDECHAR(__FUNCTION__)
The all important piece is the L##x
that appends the L character to the string constant before the compiler sees it. You can do this to __FILE__
as well using the same technique.
That looks more like a typo of __FUNCTION__
than an issue with widen()
or similar, at least if you pasted the exact error message.
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