Copy-initialization

Initializes an object from another object.

Syntax

Explanation

Copy-initialization is performed in the following situations:

The effects of copy-initialization are:

• Otherwise, 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

• Otherwise, 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 rvalue temporary(until C++11) prvalue temporary(since C++11)(until C++17) prvalue expression(since C++17) of the cv-unqualified version of T if a converting constructor was used, is then used to direct-initialize the object. The last step is usually optimized out (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

Notes

Copy-initialization is less permissive than direct-initialization: explicit constructors are not converting constructors

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

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, a move constructor

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.

Example

#include <memory>
#include <string>
#include <utility>
 
struct A
{
    operator int() { return 12;}
};
 
struct B
{
    B(int) {}
};
 
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
 
    A a;
    B b0 = 12;
//  B b1 = a;       // < error: conversion from 'A' to non-scalar type 'B' requested
    B b2{a};        // < identical, calling A::operator int(), then B::B(int)
    B b3 = {a};     // <
    auto b4 = B{a}; // <
 
//  b0 = a;         // < error, assignment operator overload needed
 
    [](...){}(c, b0, b3, b4); // pretend these variables are used
}

Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

See also

• copy elision
• converting constructor
• copy assignment
• copy constructor
• default constructor
• destructor
• explicit
• initialization
  • aggregate initialization
  • constant initialization
  • default initialization
  • direct initialization
  • initializer list
  • list initialization
  • reference initialization
  • value initialization
  • zero initialization
• move assignment
• move constructor
• new