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();
}
c++赋值构造函数为什么返回引用类型？,码迷,mamicode.com