介绍一个C++奇巧淫技
你能实现这样一个函数吗：
MyType type; HisType htype; serialize_3(11, type, htype); serialize_4(type, htype ,type, htype); serialize_4(11, type , htype, htype);
参数类型自由，个数自由，怎么做呢？往下看：
[xiaochu.yh@OB macro]$ cat auto_type.cpp /* * (C) 1999-2013 Alibaba Inc. * * This program is free software; you can redistribute it and/or modify * it under the terms of the GNU General Public License version 2 as * published by the Free Software Foundation. * * * Version: auto_type.cpp, 09/04/2013 08:02:17 PM Yu Huang Exp $ * * Author: * Huang Yu <[email protected]> * Description: * auto type match * */ #include <stdio.h> void serialize() { return; } class HisType { public: HisType(const char *i) : value_(i) { } ~HisType() { } void serialize() const { printf("HisType: s(%s)\n", value_); } private: const char *value_; }; class MyType { public: MyType(int i) : value_(i) { } ~MyType() { } void serialize() const { printf("MyType: f(%d)\n", value_); } private: int value_; }; void serialize(const int arg0) { printf("int: %d\n", arg0); } void serialize(const float arg0) { printf("float: %f\n", arg0); } template<typename Arg0> void serialize(const Arg0 &arg0) { arg0.serialize(); } template<typename Arg0> void serialize_1(const Arg0 &arg0) { serialize(arg0); } #define JOIN(x,y) JOIN2(x,y) #define JOIN2(x,y) x##y #define DECVAL_1 0 #define DECVAL_2 1 #define DECVAL_3 2 #define DECVAL_4 3 #define DEC_VAL(n) DECVAL_##n // recursively expanding macro #define ARG_TN0 #define ARG_TN1 typename Arg0 #define ARG_TN2 ARG_TN1, typename Arg1 #define ARG_TN3 ARG_TN2, typename Arg2 #define ARG_TN4 ARG_TN3, typename Arg3 #define ARG_PN0 #define ARG_PN1 const Arg0 & arg0 #define ARG_PN2 ARG_PN1, const Arg1 & arg1 #define ARG_PN3 ARG_PN2, const Arg2 & arg2 #define ARG_PN4 ARG_PN3, const Arg3 & arg3 #define ARG_AN0 #define ARG_AN1 arg0 #define ARG_AN2 ARG_AN1, arg1 #define ARG_AN3 ARG_AN2, arg2 #define ARG_AN4 ARG_AN3, arg3 #define ARG_CN0 #define ARG_CN1 arg0 #define ARG_CN2 arg1 #define ARG_CN3 arg2 #define ARG_CN4 arg3 #define SERIALIZE_DECLARE(NUM_ARG) \ template<JOIN(ARG_TN, NUM_ARG)> \ void JOIN(serialize_, NUM_ARG)(JOIN(ARG_PN, NUM_ARG)) SERIALIZE_DECLARE(2); SERIALIZE_DECLARE(3); SERIALIZE_DECLARE(4); #define SERIALIZE_DEFINE(NUM_ARG) \ template<JOIN(ARG_TN, NUM_ARG)> \ void JOIN(serialize_, NUM_ARG)(JOIN(ARG_PN, NUM_ARG)) \ { \ JOIN(serialize_, DEC_VAL(NUM_ARG))(JOIN(ARG_AN, DEC_VAL(NUM_ARG))); \ serialize(JOIN(ARG_CN, NUM_ARG)); \ } SERIALIZE_DEFINE(2); SERIALIZE_DEFINE(3); SERIALIZE_DEFINE(4); int main() { MyType type(4234); HisType htype("home"); //先来个见面礼, 1是int类型，10.2f是float类型，type是自定义类型 serialize_4(1,10.2f,3, type); printf("==============\n"); serialize_3(type,11, htype); // <== 注意下面的参数个数，以及参数顺序，完全自由！ printf("==============\n"); serialize_3(11 ,type, htype); printf("==============\n"); serialize_3(htype ,type, htype); printf("==============\n"); return 0; }
编译运行结果：
[xiaochu.yh@OB macro]$ g++ auto_type.cpp [xiaochu.yh@OB]$ ./a.out int: 1 float: 10.200000 int: 3 MyType: f(4234) ============== MyType: f(4234) int: 11 HisType: s(home) ============== int: 11 MyType: f(4234) HisType: s(home) ============== HisType: s(home) MyType: f(4234) HisType: s(home) ==============
该技术是从曲山同学的代码中学习来的，曲山对宏的运用真是炉火纯青！这里最神奇的就是下面一段代码，至今不明：
#define JOIN(x,y) JOIN2(x,y) #define JOIN2(x,y) x##y
JOIN和JOIN2不是等价的吗？不过还真不是。如果只写JOIN2，在宏展开阶段会有比较诡异的事情发生。不信你试试。但是为什么呢？我也不知道。@曲山，求助啊~~
更全面的代码见OceanBase源码oceanbase/src/common/ob_rpc_stub.h和oceanbase/src/common/ob_rpc_macros.h
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UPDATE：
这篇帖子发到了内网，得到了@探晴同学指点，加上@元启 同学的解释，基本弄明白了JOIN的机制。
原因的确很简单。
#define MY_VALUE 2
#define JOIN(A,B) A##B
JOIN(hello, world)的输出结果就是 helloworld，
JOIN(MY_VALUE, b)的输出结果就是 MY_VALUEb。尽管MY_VALUE是个宏，你期待它展开成2b。
如何成为一个2b呢？ 这么做：
#define JOIN(a, b) JOIN_EXPAND_PARAM(a,b)
#define JOIN_EXPAND_PARAM(a,b) a##b
JOIN(MY_VALUE, b)的展开过程是：
1. JOIN(MY_VALUE, b)展开成 JOIN_EXPAND_PARAM(2, b)
2. JOIN_EXPAND_PARAM(2, b) 展开成 2b
这句话：Macro arguments are completely macro-expanded before they are substituted into a macro body, unless they are stringified or pasted with other tokens.
参考： http://gcc.gnu.org/onlinedocs/cpp/Argument-Prescan.html