宏来算的参数的个数 [英] Macro to count number of arguments
问题描述
我有一个第三方的C库是一个可变参数函数:
I have a variadic function from a third-party C library:
int func(int argc, ...);
ARGC
表示传递可选参数的个数。
我与计算参数的数量,建议宏观包裹它的这里。为了方便阅读,这里的宏:
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())
和我包装它像这样:
#define my_func(...) func(PP_NARG(__VA_ARGS__), __VA_ARGS__)
的 PP_NARG
宏接受一个或多个参数的函数的伟大工程。例如, PP_NARG(你好,世界)
计算结果为 2
。结果
The PP_NARG
macro works great for functions accepting one or more arguments. For instance, PP_NARG("Hello", "World")
evaluates to 2
.
问题是,在没有参数传递, PP_NARG()
计算结果为 1
而不是 0
。结果
我明白这个宏是如何工作的,但我不能想出一个主意,使其对这种情况下的行为正确,以及对其进行修改。
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.
任何想法?
修改:结果
我已经找到了解决方法 PP_NARG
,并张贴作为一个答案。结果
我仍然有虽然包裹可变参数函数的问题。当 __ VA_ARGS __
为空, my_func,并将
扩展到 FUNC(0)
触发编译错误。
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.
推荐答案
另一种可能,不使用的sizeof
,也不是一个GCC扩展以下添加到您的$ C $ç
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
的结果是
PP_NARG() // expands to 0
PP_NARG(x) // expands to 1
PP_NARG(x, 2) // expands to 2
说明:
在这些宏的诀窍是,当与零个或一个参数,为1时,至少有两个参数调用名为 PP_HASCOMMA(...)
扩展到0。为了这两种情况之间的区别,我用 PP_COMMA __VA_ARGS__()
,当 __ VA_ARGS __
为空,并返回返回一个逗号没事的时候 __ __ VA_ARGS
非空。
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.
现在有三种可能的情况:
Now there are three possible cases:
-
__ VA_ARGS __
为空:PP_HASCOMMA(__ VA_ARGS __)
返回0和PP_HASCOMMA (PP_COMMA __VA_ARGS__())
返回1。
__VA_ARGS__
is empty:PP_HASCOMMA(__VA_ARGS__)
returns 0 andPP_HASCOMMA(PP_COMMA __VA_ARGS__ ())
returns 1.
__ VA_ARGS __
包含一个参数: PP_HASCOMMA(__ VA_ARGS __)
返回0和 PP_HASCOMMA(PP_COMMA __VA_ARGS__())
返回0。
__VA_ARGS__
contains one argument: PP_HASCOMMA(__VA_ARGS__)
returns 0 and PP_HASCOMMA(PP_COMMA __VA_ARGS__ ())
returns 0.
__ VA_ARGS __
包含两个或多个参数: PP_HASCOMMA(__ VA_ARGS __)
返回1和 PP_HASCOMMA(PP_COMMA __VA_ARGS__())
返回1。
__VA_ARGS__
contains two or more arguments: PP_HASCOMMA(__VA_ARGS__)
returns 1 and PP_HASCOMMA(PP_COMMA __VA_ARGS__ ())
returns 1.
的 PP_NARG_HELPERx
只是需要宏来解决这些案件。
The PP_NARG_HELPERx
macros are just needed to resolve these cases.
为了修复 FUNC(0)
的问题,我们需要测试我们是否已经提供了零
或多个参数。在 PP_ISZERO
宏进场这里。
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 ,
现在让我们定义这prepends参数的个数将参数列表中的另一宏:
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__
许多佣工需要再次扩大宏数值。最后测试一下:
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)
在线例如:
http://coliru.stacked-crooked.com/a/73b4b6d75d45a1c8
请也有看 P99 项目,其中有多得多
先进的preprocessor解决方案,像这些。
Please have also a look at the P99 project, which has much more advanced preprocessor solutions, like these.
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