C99中的getchar_unlocked（）隐式声明

Question

Using getchar_unlocked and compiling with --std=c99 flag gives warningas follows-

warning: implicit declaration of function 'getchar_unlocked' [-Wimplicit-function-declaration]

Does not give any warning if compiled without flag.Is there any way to work around with it ?

Answer 1

Starting from C99 you must have a visible function prototype before calling a function. While the earlier C standard would just stupidly assume that any function unknown to the compiler has the format int func (params) , which in turn would cause severe bugs most of the time.

Properly declare a prototype for getchar_unlocked and the bug will go away.

Note that there is no such function present in any standard library. It seems you might have to include some non-standard library for the compiler to find the function.

Answer 2

_unlocked versions of get... functions are POSIX extensions. They are not part of the standard functions of C99. The full list of get... functions is given in 7.19.1.5: getwc , getwchar , getc , getchar , and gets (deprecated).

When the function is not on this list, C99-compliant compiler must warn you that your program may not compile with other C99-compliant compilers.

Answer 3

Dialect selection options like -ansi and -std=c99 cause the compiler to define certain macros (in addition to altering the accepted dialect).

Library header files react to those macros.

Precisely how they react is quite system-dependent (the compiler doesn't provide a C library), but a common behavior you can broadly expect is that if you use one of these flags alone (without any other "feature selection macro"), it has the effect of hiding the declarations of functions, macros and other global symbols which are not in the specified ISO C dialect.

ISO C knows nothing about getchar_unlocked . The presence of such a declaration in <stdio.h> (normally an ISO C header) is a POSIX extension, which is basically nonconforming, since getchar_unlocked is an identifier that strictly conforming C programs can use, even if they include <stdio.h> . When you use -ansi or -std=c99 , the <stdio.h> header listens up and whips itself into ISO-C-conforming shape, hiding such extensions.

On well-behaved POSIX systems, you can request that you want an ISO C dialect and that you want certain rudimentary 1990-ish POSIX features to be visible in header files, for instance like this:

gcc -std=c99 -D_POSIX_SOURCE ...
               ^^^^^ "feature selection macro"

There is a whole science to these feature selection macros, too broad for this question and answer; some forms of them have values, like -D_XOPEN_SOURCE=500 . _POSIX_SOURCE doesn't need an argument; it is just defined or not, but _POSIX_C_SOURCE is numeric.

I just checked glibc and Cygwin: on both, _POSIX_SOURCE is enough to reveal the getchar_unlocked declaration. It is quite old, dating back to POSIX.1 1996.

Beware: on some systems, multiple feature selection macros don't play along reasonably; they give you a set intersection rather than union, so that -D_POSIX_SOURCE and -D_BSD_SOURCE together end up meaning "Declare to me only those handful of functions that are specific to classic BSD that have been standardized in POSIX too", which means that next to nothing is declared.

Answer 4

getchar_unlocked is not a C standard function.

Compiling it forcing c99 standard does not support it natively.