这是合法的避免设置创建Comparator对象的实际副本 [英] Is this legal to avoid set from creating actual copies of Comparator object

问题描述

在这样的代码中：

Comparator comp(3);

set<string, Comparator> s1(comp);
set<string, Comparator> s2(comp);
set<string, Comparator> s3(comp);
set<string, Comparator> s4(comp); 

Comparator的实际实例（即comp）在每次创建集合对象时被复制为cpp引用状态

the actual instance of the Comparator (namely comp) is copied at each creation of a set object as the cpp reference states

容器保留alloc和comp的内部副本，它们使用
分配存储并对元素。

The container keeps an internal copy of alloc and comp, which are used to allocate storage and to sort the elements throughout its lifetime.

所以我们想知道这在C ++中是否合法

So we were wondering if this is legal in C++

#include <set>
#include <iostream>

struct A {
    int i = 0;
    bool operator()(int a, int b)
    {
        ++i;
        return a < b;
    }
};

int main()
{    
    A a;
    std::set<int, A&> s1( {1, 2, 3}, a);
    std::set<int, A&> s2( {4, 5, 6}, a);
    std::cout << a.i;
}

提前感谢。

推荐答案

我无法在标准中找到禁止使用引用类型作为比较函数的措辞。因此，这似乎是合法的。注意，某些事情，如默认构造这样的集合，将被禁止，因为你的比较类型不是默认可构造的。

I'm unable to find wording in the standard forbidding using a reference type as the comparison function. Thus it seems that this would be legal. Note that some things, such as default constructing such a set, will be forbidden because your comparison type is not default constructable.

最后注意，规范的C ++方法是不是这样做，而是从外部维护状态。当你采取这种方法，它是完全清楚你在做什么，并保证是安全的：

Finally note that the canonical C++ approach is to not do this, but to maintain the state externally. When you take that approach it's totally clear what you're doing and guaranteed to be safe:

#include <set>
#include <iostream>

struct A {
    int& i_;
    explicit A(int& state) : i_(state) { }
    bool operator()(int a, int b)
    {
        ++i_;
        return a < b;
    }
};

int main() {
    int i;
    std::set<int, A> s1( {1, 2, 3}, A(i));      
    std::set<int, A> s2( {4, 5, 6}, A(i));        
    std::cout << i << endl;
}

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