Behavior of #define in c

#defi开发者_C百科ne int_p int*
int_p p1,p2,p3; // only p1 is a pointer !

can somebody exlplain why it is so.

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#define is just a textual substitution. The code above is equivalent to

int *p1, p2, p3;

so only p1 is a pointer. You need

typedef int* int_p;

instead.

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Rather than thinking of it like this:

int* (p1, p2, p3);

think of it like this:

int (*p1), p2, p3;

As in, only the symbol with the asterisk in-front of it becomes a pointer, not all of them.

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Two points:

The preprocessor just does text substitution on the source code before compilation; it has no awareness of types or syntax. After preprocessing, the lines

#define int_p int*
int_p p1, p2, p3;

expand to

int* p1, p2, p3;

Which brings us to our second point; in a declaration, the * binds to the nearest declarator, not the type specifier; IOW, the above declaration is parsed as

int (*p1), p2, p3;

Whitespace makes no difference; int* p1; is parsed the same as int *p1; as int * p1;.

If you want to declare all three variables as pointers, you have three choices:

1. Do it all by hand:

   
       int *p1, *p2, *p3;
   

2. Use your macro, but use multiple declarations

   
       int_p p1;
       int_p p2;
       int_p p3;
   

3. Create a type synonym using the `typedef` facility:

   
       typedef int *int_p; 
       int_p p1, p2, p3;
   

Unlike the preprocessor macro, the typedef is not a simple text substitution; the compiler basically creates a synonym for the type int *, and that synonym can be used anywhere int * can be used.