关于通过引用传递的怀疑

Question

我正在学习C++,特别是我已停止参考.如果我的问题对你们大多数人来说都是微不足道的,我很抱歉,但我想了解这个程序的输出:

#include <iostream>

using namespace std;

struct myStruct
{
    int a;
    int b;
};
typedef struct myStruct myStruct;

myStruct copyMyStruct(myStruct& source)
{
    myStruct dest;
    dest.a=source.a;
    dest.b=source.b;
    return dest;
}

myStruct otherCopyMyStruct(myStruct& source)
{
    myStruct dest;
    dest=source;
    return dest;
}

myStruct& GetRef(myStruct& source)
{
    return source;
}

void printMyStruct(string name,const myStruct& str)
{
    cout<<name<<".a:"<<str.a<<endl;
    cout<<name<<".b:"<<str.b<<endl;
}

myStruct one,two,three,four;
myStruct& five=one;

void printStructs()
{
    printMyStruct("one",one);
    printMyStruct("two",two);
    printMyStruct("three",three);
    printMyStruct("four",four);
    printMyStruct("five",five);
}

int main()
{
    one.a=100;
    one.b=200;

    two=copyMyStruct(one);
    three=otherCopyMyStruct(one);
    four=GetRef(one);


    printStructs();

    cout<<endl<<"NOW MODIFYING one"<<endl;

    one.a=12345;
    one.b=67890;

    printStructs();

    cout<<endl<<"NOW MODIFYING two"<<endl;

    two.a=2222;
    two.b=2222;

    printStructs();

    cout<<endl<<"NOW MODIFYING three"<<endl;

    three.a=3333;
    three.b=3333;

    printStructs();

    cout<<endl<<"NOW MODIFYING four"<<endl;

    four.a=4444;
    four.b=4444;

    printStructs();


    cout<<endl<<"NOW MODIFYING five"<<endl;

    five.a=5555;
    five.b=5555;

    printStructs();

    return 0;
}

输出是:

one.a:100
one.b:200
two.a:100
two.b:200
three.a:100
three.b:200
four.a:100
four.b:200
five.a:100
five.b:200

NOW MODIFYING one
one.a:12345
one.b:67890
two.a:100
two.b:200
three.a:100
three.b:200
four.a:100
four.b:200
five.a:12345
five.b:67890

NOW MODIFYING two
one.a:12345
one.b:67890
two.a:2222
two.b:2222
three.a:100
three.b:200
four.a:100
four.b:200
five.a:12345
five.b:67890

NOW MODIFYING three
one.a:12345
one.b:67890
two.a:2222
two.b:2222
three.a:3333
three.b:3333
four.a:100
four.b:200
five.a:12345
five.b:67890

NOW MODIFYING four
one.a:12345
one.b:67890
two.a:2222
two.b:2222
three.a:3333
three.b:3333
four.a:4444
four.b:4444
five.a:12345
five.b:67890

NOW MODIFYING five
one.a:5555
one.b:5555
two.a:2222
two.b:2222
three.a:3333
three.b:3333
four.a:4444
four.b:4444
five.a:5555
five.b:5555

我的问题:为什么"两个","三个"和"四个"的变化不会对"一个"产生变化？

我可以猜到"两个"和"三个"会发生什么:可能是成员复制到新创建的变量的成员,但我不明白为什么"四"的变化没有反映在"一个"(和"五"):毕竟我从GetRef函数返回一个引用....

提前致谢!

Answer 1

变量four是一个对象,而不是一个引用.

从引用分配给它时four=GetRef(one);,four不会变成引用.赋值复制引用引用的任何内容(在这种情况下,该引用是one).之后,对象是不相关的.所以four = GetRef(one);有同样的效果four = one;.

myStruct &five = one;另一方面,声明five为引用(不是对象),并将对象"绑定" one到引用.因此,名称five和名称one引用相同的对象,这意味着当然可以使用其他名称查看使用任一名称所做的更改.

顺便说一下,typedef struct myStruct myStruct;在C++中不需要[*] .在C++中,类类型只能通过名称引用,而结构类型是类.

[*]除了一个有点奇怪的角落情况,你有一个与类同名的函数,其参数与该类的某些构造函数的参数兼容.那么你可能会认为Foo(x,y)函数Foo和构造函数调用之间的表达式是不明确的Foo.但是没有 - 在没有typedef的情况下,C当然会选择函数,因此为了与C兼容,C++也做同样的事情.大多数人都认为这种情况不足以在C++中编写typedef.