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c++11 函数内部返回对象使用move语义的最佳实践(一)

2015-11-21 01:04:59 · 作者: · 浏览: 21
标签: 函数 内部 返回 对象 使用 move 语义 最佳 实践

一句话，直接返回即可，不用任何变化。

当启动了c++11选项后，通过函数返回代码没有发生任何变化，但是已经使用了move语义，而不需要之前的NRVO编译器优化技术。

注意，右值引用rvalue reference是表达式计算完成后就不再存在的临时变量，左值是表达式计算完成后的变量。如果能够用&求址的，就是左值。

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下面是stackoverflow上的一个讨论贴，比较有价值：

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246down voteaccepted

First example

std::vector
       
         return_vector(void) { std::vector
        
          tmp {1,2,3,4,5}; return tmp; } std::vector
         
           &&rval_ref = return_vector();

The first example returns a temporary which is caught by rval_ref. That temporary will have its life extended beyond the rval_ref definition and you can use it as if you had caught it by value. This is very similar to the following:

const std::vector
       
        & rval_ref = return_vector();

except that in my rewrite you obviously can't use rval_ref in a non-const manner.

Second example

std::vector
       
        && return_vector(void) { std::vector
        
          tmp {1,2,3,4,5}; return std::move(tmp); } std::vector
         
           &&rval_ref = return_vector();

In the second example you have created a run time error. rval_ref now holds a reference to the destructed tmp inside the function. With any luck, this code would immediately crash.

Third example

std::vector
       
         return_vector(void) { std::vector
        
          tmp {1,2,3,4,5}; return std::move(tmp); } std::vector
         
           &&rval_ref = return_vector();

Your third example is roughly equivalent to your first. The std::move on tmp is unnecessary and can actually be a performance pessimization as it will inhibit return value optimization.

The best way to code what you're doing is:

Best practice

std::vector
       
         return_vector(void) { std::vector
        
          tmp {1,2,3,4,5}; return tmp; } std::vector
         
           rval_ref = return_vector();

I.e. just as you would in C++03. tmp is implicitly treated as an rvalue in the return statement. It will either be returned via return-value-optimization (no copy, no move), or if the compiler decides it can not perform RVO, then it will use vector's move constructor to do the return. Only if RVO is not performed, and if the returned type did not have a move constructor would the copy constructor be used for the return.

shareeditflag answered Feb 13 '11 at 20:52\Howard Hinnant
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Great thanks for the elaborated answer ! – Tarantula Feb 14 '11 at 17:10
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Compilers will RVO when you return a local object by value, and the type of the local and the return of the function are the same, and neither is cv-qualified (don't return const types). Stay away from returning with the condition (:?) statement as it can inhibit RVO. Don't wrap the local in some other function that returns a reference to the local. Just return my_local;. Multiple return statements are ok and will not inhibit RVO. – Howard Hinnant Feb 25 '13 at 20:18
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There is a caveat: when returning a member of a local object, the move must be explicit. – boycy Feb 26 '13 at 8:58
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@NoSenseEtAl: There is no temporary created on the return line. move doesn't create a temporary. It casts an lvalue to an xvalue, making no copies, creating nothing, destroying nothing. That example is the exact same situation as if you returned by lvalue-reference and removed the move from the return line: Either way you've got a dangling reference to a local variable inside the function and which has been destructed. – Howard Hinnant Feb 27 '13 at 16:11
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Just a nit: Since you named the variable (tmp) in the "Best practice" section, it is the NRVO that kicks in, not the RVO. These are two different optimizations. Other than tha