Does #ifdef (or other Preprocessor Directives) Work for Function Declarations (to Test for Function Existence)?
Why doesn’t the following code work as expected?
void foobar(int);
#ifndef foobar
printf("foobar exists");
#endif
It always prints the message; it obviously cannot detect the existence of a function as an entity. (Is it an over-loading issue?)
Why can’t #ifdef
(or its vari开发者_运维技巧ants) detect function declarations? Declarations should be available at pre-processing, so it should work, shouldn’t it? If not, is there an alternative or work-around?
Declarations should be available at pre-processing, so it should work, shouldn’t it?
The pre-processor operates before the compilation (hence the "pre") so there are no compiled symbols at that point, just text and text expansion. The pre-procesor and the compiler are distinctly separate and work independantly of each other, except for the fact that the pre-processor modifies the source that is passed to the compiler.
The typical pattern to doing something like with the pre-processor is to pair the function declaration with the function usage using the same define
constant:
#define FOO
#ifdef FOO
void foo(int);
#endif
#ifdef FOO
printf("foo exists");
#endif
The pre-processor works against pre-processor tokens that are defined using pre-processor directives. The pre-processor does not work against types declared in your code (whatever the language may be).
You must use the #define preprocessor directive to declare a token visible to other pre-processor condition checks.
#define FOO
#if FOO
// this code will compile
int x = 5;
#else
// this code won't
int x = 10;
#endif
精彩评论