关于通过引用传递的怀疑

Question

i'm studying C++ and, in particular,i have stopped on references. I apologize if my question will be trivial to most of you,but i would like to understand the output of this program:

#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;
}

The output is:

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

My question: why don't the changes on "two","three" and "four" produce a change on "one"?

I can guess what happens to "two" and "three": probably a member by member copy to a newly created variable, but i don't understand why the change on "four" isn't reflected on "one" (and "five"): after all i return a reference from the GetRef function ....

Thanks in advance!

Answer 1

The variable four is an object, not a reference.

When you assign to it from a reference, four=GetRef(one); , four doesn't turn into a reference. The assignment copies whatever is referred to by the reference (in this case, that referand is one ). After that, the objects are unrelated. So four = GetRef(one); has the same effect as four = one; .

myStruct &five = one; , on the other hand, declares five as a reference (not an object), and "binds" the object one to the reference. So the name five and the name one refer to the same object, meaning that of course changes made using either name can be seen using the other name.

By the way, there's no need[*] for typedef struct myStruct myStruct; in C++. In C++, a class type can just be referred to by name, and structs are classes.

[*] except for a somewhat weird corner case where you have a function with the same name as a class, whose parameters are compatible with the parameters of some constructor of that class. Then you might think the expression Foo(x,y) would be ambiguous between the function Foo and the constructor call on Foo . But no - in the absence of a typedef, C of course chooses the function, and so for compatibility with C, C++ does the same thing. Most people don't find this case compelling enough to write the typedefs in C++.

Answer 2

Because copymystruct returns 'by value'. It involves creating a seperate temporary object which is returned by value and assigned to the result which is a separate object.