从 C++11 中的函数调用返回 std::vector 的正确方法(移动语义)

我想填充 std::vector(或其他一些 STL 容器):

I want to fill std::vector (or some other STL container):

class Foo {
public:
  Foo(int _n, const Bar &_m);
private:
  std::vector<Foo> fooes_;
}

1.好看的演员,昂贵的性能

std::vector<Foo> get_vector(int _n, const Bar &_m) {
  std::vector<Foo> ret;
  ... // filling ret depending from arguments
  return ret;
}

Foo::Foo(int _n, const Bar &_m) : fooes_(get_vector(_n, _m) {}

2.更好的性能,更糟糕的 ctor

void fill_vector(int _n, const Bar &_m, std::vector<Foo> &_ret) {
  ... // filling ret depending from arguments
}

Foo::Foo(int _n, const Bar &_m) { fill_vector(_n, _m, fooes_); }

是否可以使用 C++0x(移动语义功能等)重写第一个示例中的 get_vector 函数以避免冗余复制和构造函数调用?

Is it possible to rewrite get_vector function from 1st example with C++0x (move semantics features and so on) to avoid redundant copying and constructor calls?

推荐答案

如果您使用的是与 C++0x 兼容的编译器和标准库,您可以从第一个示例中获得更好的性能什么都不做.get_vector(_n, _m) 的返回值是一个临时值,std::vector 的移动构造函数(一个接受右值引用的构造函数)将被自动调用您无需再做进一步的工作.

If you're using a C++0x-compatible compiler and standard library, you get better performance from the first example without doing anything. The return value of get_vector(_n, _m) is a temporary, and the move constructor for std::vector (a constructor taking an rvalue reference) will automatically be called with no further work on your part.

一般来说,非库作者不需要直接使用右值引用;你会自动获得相当大的好处.

In general, non-library writers won't need to use rvalue references directly; you'll just reap a decent chunk of the benefits automatically.

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