用于计算参数数量的宏 [英] Macro to count number of arguments
问题描述
我有一个来自第三方 C 库的可变参数函数:
int func(int argc, ...);argc 表示传递的可选参数的数量.我用一个计算参数数量的宏来包装它，正如建议的 此处.为了阅读方便，这里是宏:
#define PP_ARG_N( \_1, _2, _3, _4, _5, _6, _7, _8, _9, _10, \_11, _12, _13, _14, _15, _16, _17, _18, _19, _20, \_21, _22, _23, _24, _25, _26, _27, _28, _29, _30, \_31, _32, _33, _34, _35, _36, _37, _38, _39, _40, \_41, _42, _43, _44, _45, _46, _47, _48, _49, _50, \_51, _52, _53, _54, _55, _56, _57, _58, _59, _60, \_61, _62, _63, N, ...) N#define PP_RSEQ_N() \63, 62, 61, 60, \59, 58, 57, 56, 55, 54, 53, 52, 51, 50, \49, 48, 47, 46, 45, 44, 43, 42, 41, 40, \39, 38, 37, 36, 35, 34, 33, 32, 31, 30, \29, 28, 27, 26, 25, 24, 23, 22, 21, 20, \19, 18, 17, 16, 15, 14, 13, 12, 11, 10, \9, 8, 7, 6, 5, 4, 3, 2, 1, 0#define PP_NARG_(...) PP_ARG_N(__VA_ARGS__)#define PP_NARG(...) PP_NARG_(__VA_ARGS__, PP_RSEQ_N())我是这样包装的:
#define my_func(...) func(PP_NARG(__VA_ARGS__), __VA_ARGS__)PP_NARG 宏非常适用于接受一个或多个参数的函数.例如，PP_NARG("Hello", "World") 的计算结果为 2.
问题在于，当没有传递任何参数时，PP_NARG() 的计算结果为 1 而不是 0.
我了解这个宏是如何工作，但我想不出修改它的想法，以便它在这种情况下也能正常运行.
有什么想法吗?
<小时>编辑:
我找到了 PP_NARG 的解决方法，并将其作为答案发布.
尽管如此，我仍然在包装可变参数函数时遇到问题.当 __VA_ARGS__ 为空时，my_func 扩展为 func(0, ) 这会触发编译错误.
另一种不使用 sizeof 或 GCC 扩展的可能性是将以下内容添加到您的代码中
#define PP_COMMASEQ_N() \1, 1, 1, 1, \1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \1, 1, 1, 1, 1, 1, 1, 1, 0, 0#define PP_COMMA(...) ,#define PP_HASCOMMA(...) \PP_NARG_(__VA_ARGS__, PP_COMMASEQ_N())#define PP_NARG(...) \PP_NARG_HELPER1(\PP_HASCOMMA(__VA_ARGS__), \PP_HASCOMMA(PP_COMMA __VA_ARGS__ ()), \PP_NARG_(__VA_ARGS__, PP_RSEQ_N()))#define PP_NARG_HELPER1(a, b, N) PP_NARG_HELPER2(a, b, N)#define PP_NARG_HELPER2(a, b, N) PP_NARG_HELPER3_ ## a ## b(N)#define PP_NARG_HELPER3_01(N) 0#define PP_NARG_HELPER3_00(N) 1#define PP_NARG_HELPER3_11(N) N结果是
PP_NARG()//扩展为 0PP_NARG(x)//扩展为 1PP_NARG(x, 2)//扩展为 2说明:
这些宏中的技巧是 PP_HASCOMMA(...) 在使用零个或一个参数调用时扩展为 0，在使用至少两个参数调用时扩展为 1.为了区分这两种情况，我使用了 PP_COMMA __VA_ARGS__ ()，它在 __VA_ARGS__ 为空时返回一个逗号，当 __VA_ARGS__ 为非时返回一个逗号-空.
现在有三种可能的情况:
__VA_ARGS__为空:PP_HASCOMMA(__VA_ARGS__)返回 0，PP_HASCOMMA(PP_COMMA __VA_ARGS__ ())返回 1.p>__VA_ARGS__包含一个参数:PP_HASCOMMA(__VA_ARGS__)返回 0，PP_HASCOMMA(PP_COMMA __VA_ARGS__ ())返回 0.__VA_ARGS__包含两个或更多参数:PP_HASCOMMA(__VA_ARGS__)返回 1 和PP_HASCOMMA(PP_COMMA __VA_ARGS__ ())返回 1.
PP_NARG_HELPERx 宏只需要解决这些情况.
为了修复func(0, )问题，我们需要测试我们是否提供了零或更多的论据.PP_ISZERO 宏在这里发挥作用.
#define PP_ISZERO(x) PP_HASCOMMA(PP_ISZERO_HELPER_ ## x)#define PP_ISZERO_HELPER_0 ,现在让我们定义另一个宏，该宏将参数数量添加到参数列表中:
#define PP_PREPEND_NARG(...) \PP_PREPEND_NARG_HELPER1(PP_NARG(__VA_ARGS__), __VA_ARGS__)#define PP_PREPEND_NARG_HELPER1(N, ...) \PP_PREPEND_NARG_HELPER2(PP_ISZERO(N), N, __VA_ARGS__)#define PP_PREPEND_NARG_HELPER2(z, N, ...) \PP_PREPEND_NARG_HELPER3(z, N, __VA_ARGS__)#define PP_PREPEND_NARG_HELPER3(z, N, ...) \PP_PREPEND_NARG_HELPER4_ ## z (N, __VA_ARGS__)#define PP_PREPEND_NARG_HELPER4_1(N, ...) 0#define PP_PREPEND_NARG_HELPER4_0(N, ...) N, __VA_ARGS__再次需要许多助手来将宏扩展为数值.最后测试一下:
#define my_func(...) func(PP_PREPEND_NARG(__VA_ARGS__))my_func()//扩展为 func(0)my_func(x)//扩展为 func(1, x)my_func(x, y)//扩展为 func(2, x, y)my_func(x, y, z)//扩展为 func(3, x, y, z)在线示例:
http://coliru.stacked-crooked.com/a/73b4b6d75d45a1c8
另见:
还请查看 P99 项目，其中包含多得多高级预处理器解决方案，像这些.
I have a variadic function from a third-party C library:
int func(int argc, ...);
argc indicates the number of passed optional arguments.
I'm wrapping it with a macro that counts the number of arguments, as suggested here. For reading convenience, here's the macro:
#define PP_ARG_N( \
_1, _2, _3, _4, _5, _6, _7, _8, _9, _10, \
_11, _12, _13, _14, _15, _16, _17, _18, _19, _20, \
_21, _22, _23, _24, _25, _26, _27, _28, _29, _30, \
_31, _32, _33, _34, _35, _36, _37, _38, _39, _40, \
_41, _42, _43, _44, _45, _46, _47, _48, _49, _50, \
_51, _52, _53, _54, _55, _56, _57, _58, _59, _60, \
_61, _62, _63, N, ...) N
#define PP_RSEQ_N() \
63, 62, 61, 60, \
59, 58, 57, 56, 55, 54, 53, 52, 51, 50, \
49, 48, 47, 46, 45, 44, 43, 42, 41, 40, \
39, 38, 37, 36, 35, 34, 33, 32, 31, 30, \
29, 28, 27, 26, 25, 24, 23, 22, 21, 20, \
19, 18, 17, 16, 15, 14, 13, 12, 11, 10, \
9, 8, 7, 6, 5, 4, 3, 2, 1, 0
#define PP_NARG_(...) PP_ARG_N(__VA_ARGS__)
#define PP_NARG(...) PP_NARG_(__VA_ARGS__, PP_RSEQ_N())
and I'm wrapping it like so:
#define my_func(...) func(PP_NARG(__VA_ARGS__), __VA_ARGS__)
The PP_NARG macro works great for functions accepting one or more arguments. For instance, PP_NARG("Hello", "World") evaluates to 2.
The problem is that when no arguments are passed, PP_NARG() evaluates to 1 instead of 0.
I understand how this macro works, but I can't come up with an idea to modify it so that it behaves correctly for this case as well.
Any ideas?
EDIT:
I have found a workaround for PP_NARG, and posted it as an answer.
I still have problems with wrapping the variadic function though. When __VA_ARGS__ is empty, my_func expands to func(0, ) which triggers a compilation error.
Another possibility, which does not use sizeof nor a GCC extension is to add the following to your code
#define PP_COMMASEQ_N() \
1, 1, 1, 1, \
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \
1, 1, 1, 1, 1, 1, 1, 1, 0, 0
#define PP_COMMA(...) ,
#define PP_HASCOMMA(...) \
PP_NARG_(__VA_ARGS__, PP_COMMASEQ_N())
#define PP_NARG(...) \
PP_NARG_HELPER1( \
PP_HASCOMMA(__VA_ARGS__), \
PP_HASCOMMA(PP_COMMA __VA_ARGS__ ()), \
PP_NARG_(__VA_ARGS__, PP_RSEQ_N()))
#define PP_NARG_HELPER1(a, b, N) PP_NARG_HELPER2(a, b, N)
#define PP_NARG_HELPER2(a, b, N) PP_NARG_HELPER3_ ## a ## b(N)
#define PP_NARG_HELPER3_01(N) 0
#define PP_NARG_HELPER3_00(N) 1
#define PP_NARG_HELPER3_11(N) N
The result is
PP_NARG() // expands to 0
PP_NARG(x) // expands to 1
PP_NARG(x, 2) // expands to 2
Explanation:
The trick in these macros is that PP_HASCOMMA(...) expands to 0 when called with zero or one argument and to 1 when called with at least two arguments. To distinguish between these two cases, I used PP_COMMA __VA_ARGS__ (), which returns a comma when __VA_ARGS__ is empty and returns nothing when __VA_ARGS__ is non-empty.
Now there are three possible cases:
__VA_ARGS__is empty:PP_HASCOMMA(__VA_ARGS__)returns 0 andPP_HASCOMMA(PP_COMMA __VA_ARGS__ ())returns 1.__VA_ARGS__contains one argument:PP_HASCOMMA(__VA_ARGS__)returns 0 andPP_HASCOMMA(PP_COMMA __VA_ARGS__ ())returns 0.__VA_ARGS__contains two or more arguments:PP_HASCOMMA(__VA_ARGS__)returns 1 andPP_HASCOMMA(PP_COMMA __VA_ARGS__ ())returns 1.
The PP_NARG_HELPERx macros are just needed to resolve these cases.
Edit:
In order to fix the func(0, ) problem, we need to test whether we have supplied zero
or more arguments. The PP_ISZERO macro comes into play here.
#define PP_ISZERO(x) PP_HASCOMMA(PP_ISZERO_HELPER_ ## x)
#define PP_ISZERO_HELPER_0 ,
Now let's define another macro which prepends the number of arguments to an argument list:
#define PP_PREPEND_NARG(...) \
PP_PREPEND_NARG_HELPER1(PP_NARG(__VA_ARGS__), __VA_ARGS__)
#define PP_PREPEND_NARG_HELPER1(N, ...) \
PP_PREPEND_NARG_HELPER2(PP_ISZERO(N), N, __VA_ARGS__)
#define PP_PREPEND_NARG_HELPER2(z, N, ...) \
PP_PREPEND_NARG_HELPER3(z, N, __VA_ARGS__)
#define PP_PREPEND_NARG_HELPER3(z, N, ...) \
PP_PREPEND_NARG_HELPER4_ ## z (N, __VA_ARGS__)
#define PP_PREPEND_NARG_HELPER4_1(N, ...) 0
#define PP_PREPEND_NARG_HELPER4_0(N, ...) N, __VA_ARGS__
The many helpers are again needed to expand the macros to numeric values. Finally test it:
#define my_func(...) func(PP_PREPEND_NARG(__VA_ARGS__))
my_func() // expands to func(0)
my_func(x) // expands to func(1, x)
my_func(x, y) // expands to func(2, x, y)
my_func(x, y, z) // expands to func(3, x, y, z)
Online example:
http://coliru.stacked-crooked.com/a/73b4b6d75d45a1c8
See also:
Please have also a look at the P99 project, which has much more advanced preprocessor solutions, like these.
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