clang++會自動優化函數返回代碼，無論是c++11標准還是之前的標准。下面是例子：
注意，clang++ 3.5默認編譯選項不是c++11
c++ --version Ubuntu clang version 3.5.0-4ubuntu2 (tags/RELEASE_350/final) (based on LLVM 3.5.0) Target: x86_64-pc-linux-gnu Thread model: posix
class A {
public:
A() {
cout << ctor << endl;
}
A(A const& src) {
cout << copy << endl;
}
A & operator = (A const& src) {
cout << = << endl;
return *this;
}
};
A test() {
A a;
cout << &a << endl;
return a;
}
int main(int argc, char ** argv) {
try {
A a2 = test();
cout << &a2 <
輸出為：
ctor
0x7ffd4ed29998
0x7ffd4ed29998
在test函數中一個有趣的現象是，當沒有啟動c++11編譯選項，這裡是依靠clang++編譯器的自動優化技術。當啟動了c++11選項後，代碼沒有發生任何變化，但是已經使用了move語義。
我在前幾天的程序中當用這種情況的時候返回vector，遇到了core dump，換成了vector & output參數的方式得以解決。還需要進一步追蹤，是否是clang++ 3.5的bug。
下面是stackoverflow上的一個討論貼，比較有價值：
246 down vote accepted
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.
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answered Feb 13 '11 at 20:52 Howard Hinnant
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6
Great thanks for the elaborated answer ! – Tarantula Feb 14 '11 at 17:10
10
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
3
There is a caveat: when returning a member of a local object, the move must be explicit. – boycy Feb 26 '13 at 8:58
4
@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
1
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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 that, great answer! – Daniel Frey Feb
