make_unique 完美转发

问题描述

为什么标准 C++11 库中没有 std::make_unique 函数模板?我发现

Why is there no std::make_unique function template in the standard C++11 library? I find

std::unique_ptr<SomeUserDefinedType> p(new SomeUserDefinedType(1, 2, 3));

有点啰嗦.下面的不是更好吗?

a bit verbose. Wouldn't the following be much nicer?

auto p = std::make_unique<SomeUserDefinedType>(1, 2, 3);

这很好地隐藏了 new 并且只提到了一次类型.

This hides the new nicely and only mentions the type once.

无论如何，这是我对 make_unique 实现的尝试:

Anyway, here is my attempt at an implementation of make_unique:

template<typename T, typename... Args>
std::unique_ptr<T> make_unique(Args&&... args)
{
    return std::unique_ptr<T>(new T(std::forward<Args>(args)...));
}

我花了很长时间才编译std::forward 的东西，但我不确定它是否正确.是吗?std::forward(args)... 到底是什么意思?编译器对此有何看法?

It took me quite a while to get the std::forward stuff to compile, but I'm not sure if it's correct. Is it? What exactly does std::forward<Args>(args)... mean? What does the compiler make of that?

推荐答案

C++ 标准化委员会主席 Herb Sutter 在他的 博客上写道:

Herb Sutter, chair of the C++ standardization committee, writes on his blog:

C++11 不包含 make_unique 部分是一个疏忽，而且几乎肯定会在未来添加.

That C++11 doesn’t include make_unique is partly an oversight, and it will almost certainly be added in the future.

他还给出了一个与 OP 给出的实现相同的实现.

He also gives an implementation that is identical with the one given by the OP.

std::make_unique 现在是 C++14.

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