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# copy initialization

From cppreference.com
< cpp‎ | language

Initializes an object from another object

### [edit] Syntax

T object `=` other`;` | (1) |
T object `=` `{`other`}` `;` | (2) | (until C++11)
f`(`other`)` | (3) |
`return` other`;` | (4) |
`throw` object`;` `catch (`T object`)` | (5) |
T array`[`N`] = {`other`};` | (6) |

### [edit] Explanation

Copy initialization is performed in the following situations:

1) when a named variable (automatic, static, or thread-local) of a non-reference type `T` is declared with the initializer consisting of an equals sign followed by an expression.
2) (until C++11)when a named variable of a scalar type `T` is declared with the initializer consisting of an equals sign followed by a brace-enclosed expression (Note: as of C++11, this is classified as list initialization, and narrowing conversion is not allowed).
3) when passing an argument to a function by value
4) when returning from a function that returns by value
5) when throwing or catching an exception by value
6) as part of aggregate initialization, to initialize each element for which an initializer is provided

The effects of copy initialization are:

First, if `T` is a class type and the initializer is a prvalue expression whose cv-unqualified type is the same class as `T`, the initializer expression itself, rather that a temporary materialized from it, is used to initialize the destination object: see copy elision | (since C++17)

- If `T` is a class type and the cv-unqualified version of the type of other is `T` or a class derived from `T`, the non-explicit constructors of `T` 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 cv-unqualified version of the type of other is not `T` or derived from `T`, or if `T` 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 to `T` (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 (until C++17)prvalue expression (since C++17) 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. (until C++17)

- 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 of `T`.

### [edit] Notes

Copy-initialization is less permissive than direct-initialization: explicit constructors are not converting constructors and are not considered for copy-initialization.

    struct Exp { explicit Exp(const char*) {} }; // not convertible from const char*
    Exp e1("abc");  // OK
    Exp e2 = "abc"; // Error, copy-initialization does not consider explicit constructor

    struct Imp { Imp(const char*) {} }; // convertible from const char*
    Imp i1("abc");  // OK
    Imp i2 = "abc"; // OK

In addition, the implicit conversion in copy-initialization must produce `T` directly from the initializer, while, e.g. direct-initialization expects an implicit conversion from the initializer to an argument of `T`'s constructor.

    struct S { S(std::string) {} }; // implicitly convertible from std::string
    S s("abc"); // OK: conversion from const char[4] to std::string
    S s = "abc"; // Error: no conversion from const char[4] to S
    S s = "abc"s; // OK: conversion from std::string to S

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

Run this code

    #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
    }

### [edit] See also

- default initialization
- direct initialization
- aggregate initialization
- list initialization

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