Explicit copy constructor

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The explicit copy constructor means that the copy constructor will not be called implicitly, which is what happens in the expression:

CustomString s = CustomString("test");

This expression literally means: create a temporary CustomString using the constructor that takes a const char*. Implicitly call the copy constructor of CustomString to copy from that temporary into s.

Now, if the code was correct (i.e. if the copy constructor was not explicit), the compiler would avoid the creation of the temporary and elide the copy by constructing s directly with the string literal. But the compiler must still check that the construction can be done and fails there.

You can call the copy constructor explicitly:

CustomString s( CustomString("test") );

But I would recommend that you avoid the temporary altogether and just create s with the const char*:

CustomString s( "test" );

Which is what the compiler would do anyway…

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