C 11 Move Constructor Not Called, Default Constructor Preferred
In certain scenarios, the C 11 move constructor may not be invoked, favoring the default constructor instead. To understand why, let's examine an example.
Consider the following class:
<code class="cpp">class X { public: explicit X (char* c) { cout << "ctor" << endl; init(c); }; X (X& lv) { cout << "copy" << endl; init(lv.c_); }; X (X&& rv) { cout << "move" << endl; c_ = rv.c_; rv.c_ = nullptr; }; const char* c() { return c_; }; private: void init(char *c) { c_ = new char[strlen(c)+1]; strcpy(c_, c); }; char* c_; };</code>
With this class, we can create objects as follows:
<code class="cpp">int main() { X x("test"); cout << x.c() << endl; X y(x); cout << y.c() << endl; X z( X("test") ); cout << z.c() << endl; return 0; }</code>
The expected output would be:
ctor test copy test ctor <-- Why not move? test
However, we observe that the move constructor is not called in the last line. Instead, the default constructor is used. To explain this, we need to understand copy elision.
Copy elision is an optimization technique allowed by the C 11 standard under certain conditions. It allows the compiler to construct a temporary object directly into the target object, avoiding the copy/move constructor and destructor overhead.
In this example, the X object created from the temporary "test" is elided into z. This means that no copy/move constructor is called, and instead, the object is constructed directly into z.
The compiler may perform copy elision when all of these conditions are met:
In our case, these conditions are satisfied, and thus, copy elision occurs, causing the default constructor to be used instead of the move constructor. To explicitly call the move constructor, you can use std::move as seen in:
<code class="cpp">X z( std::move(X("test")) );</code>
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