复制构造函数不调用? [英] Copy constructor not called?
问题描述
据我所知,在下列情况下调用复制构造函数:
1当实例化一个对象并使用另一个对象的值初始化它时
2当传递一个对象按值。
3当一个对象通过值从函数返回时
我决定把这个测试,我做这个小程序测试这个(每次都有消息一个构造函数被调用,它似乎适用于前两种情况,但不适用于第三种情况。我想知道我的错误,欢迎创意。
As far as I know you call the copy constructor in the following cases:
1 When instantiating one object and initializing it with values from another object
2 When passing an object by value.
3 When an object is returned from a function by value.
I decided to put this to the test and I made this small program testing this (with messages each time a constructor is called. It seems to work for the first two cases, but not for the third one. I want to find out my mistake. Ideas are welcomed.
#include <iostream>
using namespace std;
class Circle{
private:
double* data;
public:
Circle();
Circle(double* set);
Circle(const Circle& tt1);
~Circle();
Circle& operator=(const Circle& tt1);
};
Circle :: Circle()
{
cout << "Default constructor called" << endl;
data = NULL;
}
Circle :: Circle(double* set)
{
cout << "Set up constructor called" << endl;
data = new double[3];
copy(set, set+3, data);
}
Circle :: Circle(const Circle& tt1)
{
cout << "Copy constructor called" << endl;
data = new double[3];
copy(tt1.data, tt1.data+3, this->data);
}
Circle :: ~Circle()
{
cout << "Destructor called!" << endl;
delete[] data;
}
Circle& Circle :: operator=(const Circle& tt1)
{
cout << "Overloaded = called" << endl;
if(this != &tt1)
{
delete[] this->data;
this->data = new double[3];
copy(tt1.data, tt1.data+3, this->data);
}
return *this;
}
void test2(Circle a)
{
}
Circle test3()
{
double arr [] = { 3, 5, 8, 2};
Circle asd(arr);
cout<< "end of test 3 function" << endl;
return asd;
}
int main()
{
cout <<"-------------Test for initialization" << endl;
double arr [] = { 16, 2, 7};
Circle z(arr);
Circle y = z;
cout << "-------------Test for pass by value" << endl;
test2(z);
cout <<"------------- Test for return value-------"<<endl;
Circle work = test3();
cout<< "-----------Relese allocated data" << endl;
return 0;
}
推荐答案
http://en.wikipedia.org/wiki/Return_value_optimization\">返回值优化第三个测试用例的复制构造函数调用由编译器优化。
Because of Return Value Optimization the 3rd test case's copy constructor call is optimized away by the compiler.
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