如何使用初始化列表初始化不可复制的容器? [英] How to initialize a container of noncopyable with initializer list?

问题描述

可能重复:
我可以列出初始化一个只移动类型的向量吗?

我使用 gcc 4.6.1 编译这段代码

I use gcc 4.6.1 to compile this code

int main()
{
    std::vector<std::unique_ptr<int>> vec({
            std::unique_ptr<int>(new int(0)),
            std::unique_ptr<int>(new int(1)),
        });
    return 0;
}

在 g++ 的抱怨中，有类似的东西

In what g++ complains there is something like

/usr/lib/gcc/x86_64-unknown-linux-gnu/4.6.1/../../../../include/c++/4.6.1/bits/stl_construct.h:76:7: **error: use of deleted function 'std::unique_ptr<_Tp, _Dp>::unique_ptr(const std::unique_ptr<_Tp, _Dp>&)** [with _Tp = int, _Dp = std::default_delete<int>, std::unique_ptr<_Tp, _Dp> = std::unique_ptr<int>]'

在这种情况下，g++ 似乎仍在尝试复制构造函数，尽管我放入初始化列表的是 r 值.那么如何使用初始化列表初始化一个不可复制的容器呢?

It seems g++ still tries copy constructor in this case, though what I have put into initializer list are r-values. So how could I initialize a container of noncopyable with initializer list?

推荐答案

您不能将对象移出初始化列表，因为它们只允许对其成员进行 const 访问.因此，您不能使用初始化列表和移动构造函数；它们只能被复制.

You can't move objects out of initializer lists, since they only allow const access to their members. As such, you can't use initializer lists and move constructors; they can only be copied.

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