61 lines
1.3 KiB
C++
61 lines
1.3 KiB
C++
// 拷贝构造函数的调用时机
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// 编译器优化，不会调用拷贝构造函数，而是直接创建一个返回值的引用
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#include <iostream>
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using namespace std;
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class A
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{
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public:
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int x = 0;
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public:
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A()
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{
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cout << this << " A()" << endl;
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}
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A(const A &other)
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{
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this->x = other.x;
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cout << this << " A(const A &other)" << endl;
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}
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~A()
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{
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cout << this << " ~A()" << endl;
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}
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};
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void test1(A &a) // ？是否调用拷贝构造函数
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// void test1(A a) // ？是否调用拷贝构造函数
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{
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cout << "test1(A a) a.x = " << a.x << endl;
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// 当函数结束时，a 释放，会调用析构函数
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}
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A test2() // 不要声明为 A &test2(), 因为返回对象时，会释放局部对象，导致返回的引用指向一个已经释放的对象
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{
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A a1; // 局部的类对象， 没有执行拷贝构造函数
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a1.x = 1;
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cout << "test2() a1.x = " << a1.x << endl;
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return a1; // 释放局部对象 a1
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}
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void test3()
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{
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cout << "tsss = " << endl;
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A a2 = test2(); // 返回 A 类的对象，没有调用拷贝构造函数，只是创建了一个 test2() 返回值的引用
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cout << "test3() a2.x = " << a2.x << endl;
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}
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int main()
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{
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A a0;
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test1(a0); // 作为值传递，调用拷贝构造函数
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// test3();
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cout << "over" << endl;
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return 0;
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}
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