Macro to count number of arguments

Eitan T picture Eitan T · Jul 3, 2012 · Viewed 11.4k times · Source

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.

Answer

Mehrwolf picture Mehrwolf · Jul 31, 2012

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:

  1. __VA_ARGS__ is empty: PP_HASCOMMA(__VA_ARGS__) returns 0 and PP_HASCOMMA(PP_COMMA __VA_ARGS__ ()) returns 1.

  2. __VA_ARGS__ contains one argument: PP_HASCOMMA(__VA_ARGS__) returns 0 and PP_HASCOMMA(PP_COMMA __VA_ARGS__ ()) returns 0.

  3. __VA_ARGS__ contains two or more arguments: PP_HASCOMMA(__VA_ARGS__) returns 1 and PP_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.