More complex than object-like macros, a function-like macro definition declares the names of formal parameters within parentheses, separated by commas. An empty formal parameter list is legal: such a macro can be used to simulate a function that takes no arguments. C99 adds support for function-like macros with a variable number of arguments. ILE C++ supports function-like macros with a variable number of arguments, as a language extension for compatibility with C.
For portability, you should not have more than 31 parameters for a macro. The parameter list may end with an ellipsis (...). In this case, the identifier __VA_ARGS__ may appear in the replacement list.
If the identifier list does not end with an ellipsis, the number of arguments in a macro invocation must be the same as the number of parameters in the corresponding macro definition. During parameter substitution, any arguments remaining after all specified arguments have been substituted (including any separating commas) are combined into one argument called the variable argument. The variable argument will replace any occurrence of the identifier __VA_ARGS__ in the replacement list. The following example illustrates this:
#define debug(...) fprintf(stderr, __VA_ARGS__)
debug("flag"); /* Becomes fprintf(stderr, "flag"); */
Commas in the macro invocation argument list do not act as argument separators when they are:
The following line defines the macro SUM as having two parameters a and b and the replacement tokens (a + b):
#define SUM(a,b) (a + b)
This definition would cause the preprocessor to change the following statements (if the statements appear after the previous definition):
c = SUM(x,y);
c = d * SUM(x,y);
In the output of the preprocessor, these statements would appear as:
c = (x + y);
c = d * (x + y);
Use parentheses to ensure correct evaluation of replacement text. For example, the definition:
#define SQR(c) ((c) * (c))
requires parentheses around each parameter c in the definition in order to correctly evaluate an expression like:
y = SQR(a + b);
The preprocessor expands this statement to:
y = ((a + b) * (a + b));
Without parentheses in the definition, the correct order of evaluation is not preserved, and the preprocessor output is:
y = (a + b * a + b);
Arguments of the # and ## operators are converted before replacement of parameters in a function-like macro.
Once defined, a preprocessor identifier remains defined and in scope independent of the scoping rules of the language. The scope of a macro definition begins at the definition and does not end until a corresponding #undef directive is encountered. If there is no corresponding #undef directive, the scope of the macro definition lasts until the end of the translation unit.
A recursive macro is not fully expanded. For example, the definition
#define x(a,b) x(a+1,b+1) + 4
expands
x(20,10)
to
x(20+1,10+1) + 4
rather than trying to expand the macro x over and over within itself. After the macro x is expanded, it is a call to function x().
A definition is not required to specify replacement tokens. The following definition removes all instances of the token debug from subsequent lines in the current file:
#define debug
You can change the definition of a defined identifier or macro with a second preprocessor #define directive only if the second preprocessor #define directive is preceded by a preprocessor #undef directive. The #undef directive nullifies the first definition so that the same identifier can be used in a redefinition.
Within the text of the program, the preprocessor does not scan character constants or string constants for macro invocations.
The following example program contains two macro definitions and a macro invocation that refers to both of the defined macros:
/**
** This example illustrates #define directives.
**/
#include <stdio.h>
#define SQR(s) ((s) * (s))
#define PRNT(a,b) \
printf("value 1 = %d\n", a); \
printf("value 2 = %d\n", b) ;
int main(void)
{
int x = 2;
int y = 3;
PRNT(SQR(x),y);
return(0);
}
After being interpreted by the preprocessor, this program is replaced by code equivalent to the following:
#include <stdio.h>
int main(void)
{
int x = 2;
int y = 3;
printf("value 1 = %d\n", ( (x) * (x) ) );
printf("value 2 = %d\n", y);
return(0);
}
This program produces the following output:
value 1 = 4
value 2 = 3
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