C++ 用lambda代替 unique_ptr 的Deleter
C++ 用lambda代替 unique_ptr 的Deleter
代码
#include <iostream>
#include <cstdlib>
#include <memory>
#include <string>
#include <functional>
using namespace std;
class go
{
public:
go() {}
~go()
{
cout << "go die.\n";
}
};
auto d = [] ( go * gp )
{
delete gp;
cout << "deletor done.\n";
};
class go_de
{
public:
void operator() ( go* g )
{
d ( g );
}
};
int main()
{
{
unique_ptr < go, go_de > b{ new go{} };//1
}
{
//unique_ptr < go, decltype (d) > b{ new go{}}; complie error //2
unique_ptr < go, decltype (d) > a{ new go{}, d };//3
}
{
unique_ptr < go, function<void(go*) > > a{ new go{}, d };//4
//i.e. unique_ptr < go, function<void(go*) > > a{ new go{}, [](go*gp) {delete gp;cout << "deletor done.\n"; }};
}
system ( "pause" );
return 0;
}
描述
一般的,需要给一个模板的Concept参数时,都会像代码1的实现一样传入一个实现了该Concept的类型,例如go_de
就实现了unique_ptr
的模板参数Deletor
。
今天想尝试一下使用lambda表达式的类型作为模板参数传入,发现不行。原因在于
c++14 draft n4269
5.1.2 Lambda expressions
20 The closure type associated with a lambda-expression has no default constructor and a deleted copy assignment operator. It has a defaulted copy constructor and a defaulted move constructor (12.8). [ Note: These special member functions are implicitly defined as usual, and might therefore be defined as deleted. end note ]
意思就是 lambda 表达式没有默认的构造函数,operator=也被置为deleted。只有一个默认的复制构造函数和move构造函数。很显然,unique_ptr 的实现肯定是用到了Deletor Concept的默认构造函数的。所以编译不通过。这个在
unique_ptr
构造函数页写的很清楚。
- Constructs a std::unique_ptr which owns p, initializing the stored pointer with p and value-initializing the stored deleter. Requires that Deleter is DefaultConstructible and that construction does not throw an exception.2) Constructs a std::unique_ptr which owns p, initializing the stored pointer with p and value-initializing the stored deleter. Requires that Deleter is DefaultConstructible and that construction does not throw an exception.
设想unique_ptr( pointer p, d1 );
构造函数不存在,那Lambda类型就没法作为Concept传入了。
总结
- 想用Lambda表达式的类型作为Concept,使用类型推导关键字
decltype
- Lambda的类型没有default constructor、copy assignment operator.
- 写C++库的时候,如果用到模板和Concept技术,要考虑添加Concept对象做参数的类型的构造函数从而才能不限制Lambda表达式类型作为Concept传入。
毕竟,C++语言设计的原则是尽量不限制C++语言的用户的编程方式。