Question

这里还有一份副本:

std::vector<int> &f = foo();

www.un.org/spanish/ecosoc 原型

std::vector<int> foo();

Answer 1

<>代码>foo(> 系指不能受约束的暂时物体。

其中一项:

//temporary can be bound to const-reference, so this is ok
const std::vector<int> &f = foo();  //no copy takes place.

//or save a copy of temporary
std::vector<int> f = foo(); //copy takes place (may be optimized by compiler)

阅读上述评论。在第二版(非参考版本)中,汇编者可优化收益价值,避免临时复制。这还取决于<代码>foo()的实施。

参见:

Return value optimization

Answer 2

你的法典不合法。应:

std::vect或<int> const& f = foo();

或

std::vect或<int> f = foo();

The standard allows a copy in both cases (at least C++03 did—I think C++11 tighted things up in the first case); it requires it in neither, and I don t know of a compiler which actually does a copy in either case. If you re really w或ried about the copy, though, changing the declaration of foo to:

void foo( std::vect或<int>& results );

使用:

std::vect或<int> f;
foo( f );

guarantees that there will be no copies, anywhere. It s not as pretty, but if profiling shows that copying is a bottleneck here, it should be considered.

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