What does the following statement mean:
#define FAHAD
I am familiar with the statements like:
#define FAHAD 1
But what does the #define statement without a token signify? Is it that it is similar to a constant definition?
Defining a constant without a value acts as a flag to the preprocessor, and can be used like so:
#define MY_FLAG
#ifdef MY_FLAG
/* If we defined MY_FLAG, we want this to be compiled */
#else
/* We did not define MY_FLAG, we want this to be compiled instead */
#endif
it means that FAHAD
is defined, you can later check if it's defined or not with:
#ifdef FAHAD
//do something
#else
//something else
#endif
Or:
#ifndef FAHAD //if not defined
//do something
#endif
A real life example use is to check if a function or a header is available for your platform, usually a build system will define macros to indicate that some functions or headers exist before actually compiling, for example this checks if signal.h
is available:
#ifdef HAVE_SIGNAL_H
# include <signal.h>
#endif/*HAVE_SIGNAL_H*/
This checks if some function is available
#ifdef HAVE_SOME_FUNCTION
//use this function
#else
//else use another one
#endif
Any #define
results in replacing the original identifier with the replacement tokens. If there are no replacement tokens, the replacement is empty:
#define DEF_A "some stuff"
#define DEF_B 42
#define DEF_C
printf("%s is %d\n", DEF_A, DEF_B DEF_C);
expands to:
printf("%s is %d\n", "some stuff", 42 );
I put a space between 42
and )
to indicate the "nothing" that DEF_C
expanded-to, but in terms of the language at least, the output of the preprocessor is merely a stream of tokens. (Actual compilers generally let you see the preprocessor output. Whether there will be any white-space here depends on the actual preprocessor. For GNU cpp, there is one.)
As in the other answers so far, you can use #ifdef
to test whether an identifier has been #define
d. You can also write:
#if defined(DEF_C)
for instance. These tests are positive (i.e., the identifier is defined) even if the expansion is empty.
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