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copy initialization

 

 

 

 

Initializes an object from another object

Syntax

T object = other ;	(1)
f(other);	(2)
return other;	(3)
catch ( T other) ;	(4)
T array [ N ] = { other };	(5)

Explanation

Copy initialization is performed in the following situations:

1) when a named variable (automatic, static, or thread-local) is declared with the initializer consisting of an equals sign followed by an expression.

2) when passing an argument to a function by value

3) when returning from a function that returns by value

4) when catching an exception by value

5) as part of  , to initialize each element for which an initializer is provided

The effects of copy initialization are:

• If T is a class type and the type of other is cv-unqualified version of T or a class derived from T, the constructors of Tare examined and the best match is selected by overload resolution. The constructor is then called to initialize the object.

• If T is a class type, and the type of other is different, or if T is non-class type, but the type of other is a class type, that can convert from the type of other to T are examined and the best one is selected through overload resolution. The result of the conversion, which is a prvalue temporary of the destination type, is then used to  the object. The last step is usually  and the result of the conversion function is constructed directly in the memory allocated for the target object, but the copy constructor is required to be accessible even though it's not used.

• Otherwise (if neither T nor the type of other are class types),  are used, if necessary, to convert the value of other to the cv-unqualified version of T.

Notes

Copy-initialization is less permissive than direct-initialization: copy-initialization only considers non-explicit constructors and user-defined conversion functions.

If other is an rvalue expression,  will be selected by overload resolution and called during copy-initialization.

 is defined in terms of copy-initialization: if an object of type T can be copy-initialized with expressionE, then E is implicitly convertible to T.

The equals sign, =, in copy-initialization of a named variable is not related to the assignment operator. Assignment operator overloads have no effect on copy-initialization.

 

1 #include 
       
         2 #include 
        
          3 #include 
         
           4   5 int main() 6 { 7     std::string s = "test"; // OK: constructor is non-explicit 8     std::string s2 = std::move(s); // this copy-initialization performs a move 9  10 //  std::unique_ptr
          
            p = new int(1); // error: constructor is explicit11     std::unique_ptr
           
             p(new int(1)); // OK: direct-initialization12  13     int n = 3.14;    // floating-integral conversion14     const int b = n; // const doesn't matter15     int c = b;       // ...either way16 }
           
          
         
        
       

See also