[英]Copy constructors and Assignment Operators
I wrote the following program to test when the copy constructor is called and when the assignment operator is called: 我编写了以下程序来测试何时调用复制构造函数以及何时调用赋值运算符:
#include
class Test
{
public:
Test() :
iItem (0)
{
std::cout << "This is the default ctor" << std::endl;
}
Test (const Test& t) :
iItem (t.iItem)
{
std::cout << "This is the copy ctor" << std::endl;
}
~Test()
{
std::cout << "This is the dtor" << std::endl;
}
const Test& operator=(const Test& t)
{
iItem = t.iItem;
std::cout << "This is the assignment operator" << std::endl;
return *this;
}
private:
int iItem;
};
int main()
{
{
Test t1;
Test t2 = t1;
}
{
Test t1;
Test t2 (t1);
}
{
Test t1;
Test t2;
t2 = t1;
}
}
This results in the following output (just added empy lines to make it more understandable): 这将导致以下输出(仅添加了empy行以使其更易于理解):
doronw@DW01:~$ ./test This is the default ctor This is the copy ctor This is the dtor This is the dtor This is the default ctor This is the copy ctor This is the dtor This is the dtor This is the default ctor This is the default ctor This is the assignment operator This is the dtor This is the dtor
The second and third set behave as expected, but in the first set the copy constructor is called even though the assignment operator is used. 第二和第三组的行为符合预期,但是在第一组中,即使使用了赋值运算符,也会调用复制构造函数。
Is this behaviour part of the C++ standard or just a clever compiler optimization (I am using gcc 4.4.1) 此行为是C ++标准的一部分,还是仅仅是聪明的编译器优化(我正在使用gcc 4.4.1)
No assignment operator is used in the first test-case. 在第一个测试用例中不使用赋值运算符。 It just uses the initialization form called "copy initialization".
它仅使用称为“复制初始化”的初始化形式。 Copy initialization does not consider explicit constructors when initializing the object.
初始化对象时,复制初始化不考虑显式构造函数。
struct A {
A();
// explicit copy constructor
explicit A(A const&);
// explicit constructor
explicit A(int);
// non-explicit "converting" constructor
A(char const*c);
};
A a;
A b = a; // fail
A b1(a); // succeeds, "direct initialization"
A c = 1; // fail, no converting constructor found
A d(1); // succeeds
A e = "hello"; // succeeds, converting constructor used
Copy initialization is used in those cases that correspond to implicit conversions, where one does not explicitly kick off a conversion, as in function argument passing, and returning from a function. 在与隐式转换相对应的情况下使用复制初始化,在这种情况下,隐式转换不会像函数参数传递和从函数返回那样显式启动转换。
C++ standard 8.5/12 C ++标准8.5 / 12
The initialization that occurs in argument passing, function return, throwing an exception (15.1), handling an exception (15.3), and brace-enclosed initializer lists (8.5.1) is called copy-initialization and is equivalent to the form
在参数传递,函数返回,引发异常(15.1),处理异常(15.3)和括号括起来的初始化列表(8.5.1)中发生的初始化称为复制初始化,并且等效于以下形式
T x = a;
The initialization that occurs in new expressions (5.3.4), static_cast expressions (5.2.9), functional notation type conversions (5.2.3), and base and member initializers (12.6.2) is called direct-initialization and is equivalent to the form
在新表达式(5.3.4),static_cast表达式(5.2.9),功能符号类型转换(5.2.3)以及基和成员初始化程序(12.6.2)中发生的初始化称为直接初始化,并且等效于表格
T x(a);
Your first set is according to the C++ standard, and not due to some optimization. 您的第一组内容是根据C ++标准制定的,而不是由于某些优化而造成的。
Section 12.8 ( [class.copy]
) of the C++ standard gives a similar example: C ++标准的第12.8节(
[class.copy]
)提供了一个类似的示例:
class X {
// ...
public:
X(int);
X(const X&, int = 1);
};
X a(1); // calls X(int);
X b(a, 0); // calls X(const X&, int);
X c = b; // calls X(const X&, int);
The last line would be the one matching your case. 最后一行将是与您的情况匹配的那一行。
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