Home >Backend Development >C++ >Why Doesn't My C 11 Move Constructor Get Called in This Case?
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>
And its usage:
<code class="cpp">X x("test"); cout << x.c() << endl; X y(x); cout << y.c() << endl; X z( X("test") ); cout << z.c() << endl;</code>
The output is:
ctor test copy test ctor <-- why not move? test
In VS2010, using default settings, one would expect the last object (z) to be move-constructed instead of default-constructed. Using X z( move(X("test")) ) results in the expected output: ctor move test. Could this be a case of NRVO?
Should the move constructor be called according to the C 11 standard? If so, why isn't it called?
The observed behavior is due to copy elision. The compiler can directly construct a temporary into a target it is to be copied/moved into, thus omitting the copy/move constructor and destructor calls.
The situations where copy elision can be applied are outlined in §12.8.32 of the C 11 standard:
In this case, the compiler is able to elide the copy or move operation between the temporary X("test") and the target z, resulting in the observed behavior.
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