无需复制和粘贴即可重用嵌套循环

2022-01-11 00:00:00 loops macros nested-loops foreach c++

假设我有这个嵌套循环

for (int a=1; a<MAX_A; ++a)
  for (int b=1; b<MAX_B; ++b)
    for (int c=1; c<MAX_C; ++c)
    {
       do_something(a, b ,c);
    }

我在我的代码的各个部分重用了这个循环，改变了函数 do_something.每次前三行都重写很无聊.例如，在 python 中，我会创建一个生成器来返回一个迭代器 (1, 1, 1), (1, 1, 2), ... 或类似 itertools.product.


and I reuse this loop in various part of my code, changing the function do_something. It's quite boring to rewrite every time the first three lines. In python for example I would created a generator to return an iterator (1, 1, 1), (1, 1, 2), ... or something like itertools.product.
在 c++ 中，我想到的唯一解决方案是定义一个宏.更好的东西?e

In c++ the only solution I've in mind is to define a macro. Something better?e
推荐答案
使用模板:
template<typename Func>
inline void do_something_loop(Func f)
{
    for (int a=1; a<MAX_A; ++a) 
      for (int b=1; b<MAX_B; ++b) 
        for (int c=1; c<MAX_C; ++c) 
        { 
           f(a, b ,c); 
        } 
}

可以使用任何匹配签名的函数指针或函数对象调用，例如:

This can be called with any function pointer or function object that matches the signature, e.g.:
void do_something(int a, int b, int c) { /* stuff */ }

do_something_loop(do_something);

或者用函数对象:
struct do_something
{
    void operator()(int a, int b, int c) { /* stuff */ }
};

do_something_loop(do_something()); 

或者，如果您的编译器支持 C++11，即使使用 lambda:

Or if your compiler supports C++11, even with a lambda:
do_something_loop([](int a, int b, int c) { /* stuff */ });

请注意，您也可以将 f 参数声明为带有签名 void(*f)(int,int,int) 的函数指针，而不是使用模板, 但这不太灵活(它不适用于函数对象(包括 std::bind 的结果)或 lambdas).

Note that you could also declare the f parameter as a function pointer with the signature void(*f)(int,int,int) instead of using a template, but that's less flexible (it won't work on function objects (including the result of std::bind) or lambdas).


	
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