问题
What\'s the trick to create a variadic macro FOO(a1, a2, a3,..., an) such that it expands to FOOn(a1, a2, a3,..., an) for values of n in whatever preselected bounded range you choose? That is, FOO(a) should expand to FOO1(a), FOO(a, b, c) to FOO3(a, b, c), etc. I know there\'s a standard trick but I can\'t seem to find it.
Please feel free to mark this question as a duplicate and close it if there\'s another question with the answer. I suspect there is but I couldn\'t find it.
回答1:
This post Variadic macro to count number of arguments has what you're looking for I believe. Look at the first and second responses.
回答2:
#define VA_NARGS_IMPL(_1, _2, _3, _4, _5, N, ...) N
#define VA_NARGS(...) VA_NARGS_IMPL(__VA_ARGS__, 5, 4, 3, 2, 1)
#define FOO_IMPL2(count, ...) FOO ## count (__VA_ARGS__)
#define FOO_IMPL(count, ...) FOO_IMPL2(count, __VA_ARGS__)
#define FOO(...) FOO_IMPL(VA_NARGS(__VA_ARGS__), __VA_ARGS__)
FOO(a)
FOO(a, b)
FOO(a, b, c)
The invocations are replaced by:
FOO1 (a)
FOO2 (a, b)
FOO3 (a, b, c)
回答3:
Improving upon James answer to add some flexibility:
#define VA_NARGS_IMPL(_1, _2, _3, _4, _5, N, ...) N
#define VA_NARGS(...) VA_NARGS_IMPL(X,##__VA_ARGS__, 4, 3, 2, 1, 0)
#define VARARG_IMPL2(base, count, ...) base##count(__VA_ARGS__)
#define VARARG_IMPL(base, count, ...) VARARG_IMPL2(base, count, __VA_ARGS__)
#define VARARG(base, ...) VARARG_IMPL(base, VA_NARGS(__VA_ARGS__), __VA_ARGS__)
#define MyMacro0() Also works without arguments.
#define MyMacro2(x,y) [x...y]
#define MyMacro(...) VARARG(MyMacro, __VA_ARGS__)
MyMacro()
MyMacro(a)
MyMacro(a, b)
MyMacro(a, b, c)
Output:
Also works without arguments.
MyMacro1(a)
[a...b]
MyMacro3(a, b, c)
来源：https://stackoverflow.com/questions/5365440/variadic-macro-trick