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c++赋值构造函数为什么返回引用类型?
目录
-
0 前言
-
1 内置类型
-
2 自定义类型
-
3 结论
-
4 源码
0. 前言
c++默认赋值构造函数的返回值是引用类型,c++赋值运算符=的本意是返回左值的引用,我们重载赋值构造函数的时候,返回值是否应该设为引用类型呢? 按照《Effective C++》中第10条,最好是设为引用类型。
本文,通过实验来表述返回值是否为引用类型的区别。
1. 内置类型
int i= 1, j=2, k=3;
- case1: k = j = i
i == 1
j == 1
k == 1
- case2: (k = j) = i
i == 1
j == 2
k == 1
2. 自定义类型
People p1("p1"), p2("p2"), p3("p3");
- case1: p3 = p2 = p1
-
- 使用默认赋值构造函数
p3 == p1
p2 == p1
-
- 返回引用的People
p3 == p1
p2 == p1
-
- 不返回引用的People
p3 == p1
p2 == p1
- case2: (p3 = p2) = p1
-
- 使用默认赋值构造函数
p3 == p1
p2 == p2
-
- 返回引用的People
p3 == p1
p2 == p2
-
- 不返回引用的People
p3 == p2
p2 == p2
3. 结论
- case1,是否返回引用没有影响;
- case2,是否返回引用是有区别的,由于c++内置类型的赋值重载操作符是返回引用的,所以我们应该遵循规则,类的赋值构造函数返回引用类型。
4. 实验源码
#include <algorithm> #include <cstdio> #include <cstdlib> #include <cwchar> #include <functional> #include <iostream> #include <iomanip> #include <iterator> #include <string> #include <vector> #include <memory> #include <sstream> #include <utility> using std::cout; using std::endl; using std::string; using std::vector; using std::stringstream; void test_int() { cout << "test_int()" << endl; int i = 1, j = 2, k = 3; k = j = i; cout << "\ti=" << i << endl; // 1 cout << "\tj=" << j << endl; // 1 cout << "\tk=" << k << endl; // 1 } void test_int2() { cout << "test_int2()" << endl; int i = 1, j = 2, k = 3; (k = j) = i; cout << "\ti=" << i << endl; // 1 cout << "\tj=" << j << endl; // 2 cout << "\tk=" << k << endl; // 1 } class People { public: People(const string &_name = "") : name(_name) {} People operator=(const People &_p) { name = _p.name; return *this; } string name; }; void test() { cout << "test(): not reference" << endl; cout << "\tp3=p2=p1" << endl; People p1("p1"), p2("p2"), p3("p3"); p3 = p2 = p1; cout << "\t\tp2.name=" << p2.name << endl; // p1 cout << "\t\tp3.name=" << p3.name << endl; // p1 } void test2() { cout << "test2(): not reference" << endl; cout << "\t(p3=p2)=p1" << endl; People p1("p1"), p2("p2"), p3("p3"); (p3 = p2) = p1; cout << "\t\tp2.name=" << p2.name << endl; // p2 cout << "\t\tp3.name=" << p3.name << endl; // p2 } class PeopleRef { public: PeopleRef(const string &_name = "") : name(_name) {} PeopleRef& operator=(const PeopleRef &_p) { name = _p.name; return *this; } string name; }; void test_ref() { cout << endl; cout << "test_ref(): reference" << endl; cout << "\tp3=p2=p1" << endl; PeopleRef p1("p1"), p2("p2"), p3("p3");; p3 = p2 = p1; cout << "\t\tp2.name=" << p2.name << endl; // p1 cout << "\t\tp3.name=" << p3.name << endl; // p1 } void test_ref2() { cout << "test_ref2(): reference" << endl; cout << "\t(p3=p2)=p1" << endl; PeopleRef p1("p1"), p2("p2"), p3("p3");; (p3 = p2) = p1; cout << "\t\tp2.name=" << p2.name << endl; // p2 cout << "\t\tp3.name=" << p3.name << endl; // p1 } class PeopleDefault { public: PeopleDefault(const string &_name = "") : name(_name) {} string name; }; void test_default() { cout << endl; cout << "test_default()" << endl; cout << "\tp3=p2=p1" << endl; PeopleDefault p1("p1"), p2("p2"), p3("p3"); p3 = p2 = p1; cout << "\t\tp2.name=" << p2.name << endl; // p1 cout << "\t\tp3.name=" << p3.name << endl; // p1 } void test_default2() { cout << "test_default2()" << endl; cout << "\t(p3=p2)=p1" << endl; PeopleDefault p1("p1"), p2("p2"), p3("p3"); (p3 = p2) = p1; cout << "\t\tp2.name=" << p2.name << endl; // p2 cout << "\t\tp3.name=" << p3.name << endl; // p1 } int main () { test_int(); test_int2(); test(); test2(); test_ref(); test_ref2(); test_default(); test_default2(); }
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