Is it possible to add default arguments before variable argument in variadic macro? e.g I have the version of macro something like
#define MACRO(arg1, ...) func(arg1, ##__VA_ARGS__)
I would like to add 2 more default arguments in the macro before variable arguments so that it should not affect previous version. Like:
#define MACRO(arg1, arg2 = "", arg3 = "", ...) func(arg1, arg2, arg3, ##__VA_ARGS__)
Any help would be appreciated.
I do not think this is possible. How would the compiler/preprocessor knows if the second and third arguments are part of the variable arguments or override the default values ? That's the reason why parameters with default values have to be in last position in a function.
I'm afraid you'll have to have two macros or three if you to be able to specify arg2 and use arg3 default value, but this is error-prone.
#define MACRO(arg1, ...) func(arg1, "", "", ##__VA_ARGS__)
#define MACRO2(arg1, arg2, ...) func(arg1, arg2, "", ##__VA_ARGS__)
#define MACRO3(arg1, arg2, arg3, ...) func(arg1, arg2, arg3, ##__VA_ARGS__)
What you can do:
struct foo {
int aaa;
char bbb;
char *ccc;
};
#define BAR(...) bar((struct foo){__VA_ARGS__})
void bar(struct foo baz)
/* set defaults */
if (!baz.aaa)
baz.aaa = 10;
if (!baz.bbb)
baz.bbb = 'z';
if (!baz.ccc)
baz.ccc = "default";
printf("%d, %c, %s\n", baz.aaa, baz.bbb, baz.ccc);
}
...
BAR(); // prints "10, z, default"
BAR(5); // prints "5, z, default"
BAR(5,'b'); // prints "5, b, default"
BAR(5, 'b', "something"); // prints "5, b, something"
Bad thing about this - zero parameter is treated like no parameter, e.g. BAR(0, 'c')
will produce string 10, c, default
Not as an answer to your question but as a way to simply solve your problem:
#define MACRO(arg1, ...) \
/* pre-treatment */ \
func(arg1, ##__VA_ARGS__) \
/* post-treatment */ \
void func (type1 arg1, type2 arg2 = "", type3 arg3 = "", ...);
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