copy initialization
Initializes an object from another object
Contents |
[edit] Syntax
T object = other ;
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(1) | ||||||||
f( other) ;
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(2) | ||||||||
return other;
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(3) | ||||||||
catch ( T other) ;
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(4) | ||||||||
T array [ N ] = { other };
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(5) | ||||||||
[edit] Explanation
Copy initialization is performed in the following situations:
T
is declared with the initializer consisting of an equals sign followed by an expression.The effects of copy initialization are:
- If
T
is a class type and the type of other is cv-unqualified version ofT
or a class derived fromT
, the constructors ofT
are 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 ifT
is non-class type, but the type of other is a class type, user-defined conversion sequences that can convert from the type of other toT
(or to a type derived from T if T is a class type and a conversion function is available) are examined and the best one is selected through overload resolution. The result of the conversion, which is a prvalue temporary if a converting constructor was used, is then used to direct-initialize the object. The last step is usually optimized out and the result of the conversion is constructed directly in the memory allocated for the target object, but the appropriate constructor (move or copy) is required to be accessible even though it's not used.
- Otherwise (if neither
T
nor the type of other are class types), standard conversions are used, if necessary, to convert the value of other to the cv-unqualified version ofT
.
[edit] 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, move constructor will be selected by overload resolution and called during copy-initialization. There is no such term as move-initialization.
Implicit conversion is defined in terms of copy-initialization: if an object of type T
can be copy-initialized with expression E
, 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.
[edit] Example
#include <string> #include <utility> #include <memory> int main() { std::string s = "test"; // OK: constructor is non-explicit std::string s2 = std::move(s); // this copy-initialization performs a move // std::unique_ptr<int> p = new int(1); // error: constructor is explicit std::unique_ptr<int> p(new int(1)); // OK: direct-initialization int n = 3.14; // floating-integral conversion const int b = n; // const doesn't matter int c = b; // ...either way }