std :: vector :: erase（item）需要为item定义操作符？ [英] std::vector::erase(item) needs assignment operator to be defined for item?

问题描述

我有一个类 C ，它没有定义 operator = 。我试图使用像这样的向量： std :: vector< std :: pair< C，D> vec; 。现在，我的问题是，我不能擦除这对后，我完成它。编译器抱怨缺少 operator = for C 。我不能有一个没有这个运算符的类的向量？如何解决这个问题？我无法向 C 添加分配。这是我得到的错误：

I have a class C which does not define operator=. I have am trying to use a vector like so: std::vector<std::pair<C,D>> vec;. Now, my problem is that I am not able to erase the pair after I am done with it. The compiler complains of missing operator= for C. Can I not have a vector of a class which does not have this operator? How do I get around this? I cannot add assignment to C. This is the error I get:

error C2582: 'operator =' function is unavailable in 'C'    C:\...\include\utility  196 1   my-lib

这是我的代码：

void Remove(const C& c) 
{
    auto i = cs_.begin();
    while( i != cs_.end()) {
        if (i->first == c) {
            cs_.erase(i);  // this is the problem
            break;
        }
        i++;
    }
 }

其中 cs _ 是：

std::vector<std::pair<C,D>> cs_;

推荐答案

原因是擦除会重新分配对象擦除不同于 std :: vector :: end（）的位置。重新分配涉及复制。

The reason is that erase will reallocate your objects if you erase a different position than std::vector::end(). A reallocation implicates copying.

请注意，不可复制类型的向量只能部分使用。在我们有 emplace（）（pre-C ++ 11）之前，这是不可能的。如果你的类是可复制的，为什么不定义一个赋值运算符呢？

Notice that a vector of an uncopyable type is only partially useable. Before we had emplace() (pre-C++11) it was even impossible. If your class is however copyable, why dont you define an assignment operator then?

一个解决方法可以是一个smartpointer（或正常指针） > std :: vector< std :: unique_ptr< C>>

A workaround could be a vector of smartpointers (or normal pointers) like std::vector<std::unique_ptr<C>>

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