On Thu, 11 Apr 2013 06:41:18 -0700, paul Fultz
How would I go about doing the following transformation:
(a)(b)(c)(d) -> (a, b)(c, d)
Although, you probably could use BOOST_PP_SEQ_FOR_EACH, the easiest way is to use sequence iteration instead:
#define ADJACENT_SEQ_TO_TUPLE(seq) BOOST_PP_CAT(ADJACENT_SEQ_TO_TUPLE_1 seq, _END) #define ADJACENT_SEQ_TO_TUPLE_0(x) x ADJACENT_SEQ_TO_TUPLE_2 #define ADJACENT_SEQ_TO_TUPLE_1(x) , x ADJACENT_SEQ_TO_TUPLE_2 #define ADJACENT_SEQ_TO_TUPLE_2(x) x ADJACENT_SEQ_TO_TUPLE_1 #define ADJACENT_SEQ_TO_TUPLE_0_END #define ADJACENT_SEQ_TO_TUPLE_1_END #define ADJACENT_SEQ_TO_TUPLE_2_END
Ah, thanks. I was going down the BOOST_PP_SEQ_FOR_EACH path. My algorithm was going to be #define SEQ (a)(b)(c)(d) #define WRAP(r, data, elem) \ LPAREN_IF(MOD(elem-index,2)) elem COMMA_IF(MOD(elem-index,2)) RPAREN_IF(NOT(MOD(elem-index,2))) SEQ_FOR_EACH(WRAP, ~, SEQ) But I couldn't figure out how to get the element index. BTW, what is the "next available BOOST_PP_FOR repetition" in the ternary macro for SEQ_FOR_EACH? For the example, why the is the output of: #define SEQ (w)(x)(y)(z)(a)(b)(c) #define MACRO(r, data, elem) r elem BOOST_PP_SEQ_FOR_EACH(MACRO, 1, SEQ) this: 2 w 3 x 4 y 5 z 6 a 7 b 8 c It seems r = elem-index + 2. Is this true? Thanks, Mostafa