递归typedef函数定义:std :: function返回自己的类型 [英] Recursive typedef function definition : std::function returning its own type
问题描述
我想实现一个状态机。该状态由类型 callback_t
: callback_t(int&)
的函数表示,该函数返回相同类型的函数。
I am trying to implement a state-machine. The state is represented by a function of type callback_t
: callback_t(int&)
which returns a function of same type.
我不知道如何实现它,因为递归类型函数似乎不允许。
I dont know how to implement it since recursive typed function seems not to be allowed.
尝试(作为玩具):
#include <stdio.h>
#include <functional>
typedef std::function< callback_t(int &) > callback_t ;
callback_t f1(int & i)
{
i++;
return f1;
}
callback_t f0(int & i)
{
if(i==0) i++;
return f1;
}
callback_t start(int & i)
{
i=0;
return f0;
}
int main(int argc, char **argv)
{
callback_t begin = start;
int i=0;
while(i<100)
begin = begin(i);
printf("hello world\n");
return 0;
}
错误:
C:/work/tests/tests/main.cpp:4:41: error: 'callback_t' was not declared in this scope
typedef std::function< callback_t(int &) > callback_t ;
^
有没有办法实现这种行为?
Is there a way to implement this kind of behaviour ?
Env:win7,codelite,mingw 4.8.1
Env : win7, codelite, mingw 4.8.1
推荐答案
由于递归类型定义不可能,你可以声明一个结构,携带函数并隐式转换为它:
Since recursive type definition is not possible, you can declare a structure that carry the function and implicitly cast to it:
template< typename... T >
struct RecursiveHelper
{
typedef std::function< RecursiveHelper(T...) > type;
RecursiveHelper( type f ) : func(f) {}
operator type () { return func; }
type func;
};
typedef RecursiveHelper<int&>::type callback_t;
例如: http://coliru.stacked-crooked.com/a/c6d6c29f1718e121
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