Question about multiple objects with the same name

**shiroaisu** · 06-06-2010

Lets say i have this code:

Code:

#include <iostream>

using namespace std;

class test
{
    int a;
    public:
    test(int);
    int r_a();
};

int test::r_a()
{
    return a;
}

test::test(int x)
{
    a=x;
}

int main()
{
    int y=0;
    char c;

    test b(y);
    y++;

    switch(c=cin.get())
    {
        case '1': {
            test b(y);
            y++;
        }
        default: y++; break;
    }

    cout <<b.r_a();
}

My question is:
is it possible to have more than 1 object with the same name??

and if yes, how do we distinguish them and call them individualy?? (how do u make them unique??)

If not then what happens when i press 1 and create another object with the same class and the same name??? is the current object replaced by that new one?? (for example the first object had y==0 and now as i created a new one that value is overwriten staying only 1 object with y==1),
or does the code ignore the creation of the new object assuming that we already have an object with the same name???

Sory if it is confuse, help pls.

**laserlight** · 06-06-2010

Originally Posted by shiroaisu

is it possible to have more than 1 object with the same name??

and if yes, how do we distinguish them and call them individualy?? (how do u make them unique??)

In a way yes, in different scopes or different namespaces.

Originally Posted by shiroaisu

If not then what happens when i press 1 and create another object with the same class and the same name??? is the current object replaced by that new one??

The variable named b in the scope of the switch hides the variable named b at function scope.

**VirtualAce** · 06-06-2010

The variable named b in the scope of the switch hides the variable named b at function scope.

Indeed it does and this kind of behavior should be avoided b/c if you get into the habit of doing this it can create some very nasty hard to find bugs later. You should really try to avoid any kind of name collisions. The only time it really becomes a problem is in large code lines where many people are working on the code. Thankfully you can use namespaces to avoid this kind of issue. The only time namespaces will not help is when a symbol has been defined in the global scope such as in windows.h. No amount of trickery can get past this. If you want to see an example of it then create a class and put a function called GetObject() in it and make sure you include windows.h. You will see quickly that GetObject() or GetObjectW() has been defined and your code will not compile even if it is in a class in a namespace.

**laserlight** · 06-06-2010

Originally Posted by Bubba

The only time namespaces will not help is when a symbol has been defined in the global scope such as in windows.h. No amount of trickery can get past this.

The problem arises when the name is a macro name, since macros do not obey the rules of scope. Otherwise, one can have a name in the global namespace be the same as a name in some other namespace.

**shiroaisu** · 06-06-2010

The variable named b in the scope of the switch hides the variable named b at function scope.

sory i didnt really got this, ur saying that the second object called "g" hides the first one?? and if yes what do u mean by hiding???

**laserlight** · 06-06-2010

Originally Posted by shiroaisu

sory i didnt really got this, ur saying that the second object called "g" hides the first one?? and if yes what do u mean by hiding???

There is no variable named g, but you probably mean b, and if so, yes. Here, hiding means that the name b in the scope of the switch refers to a different object from the name b at function scope. Effectively, you are unable to use the name b to access that other object, because that variable named b is hidden.

**iMalc** · 06-06-2010

To demonstrate variable hiding to yourself, run the following function, or at least tell us what you think it will print:

Code:

int foobar()
{
    int foo = 1;
    cout << foo;
    {
        int foo = 2;
        cout << foo;
        {
            int foo = 3;
            cout << foo;
        }
        cout << foo;
    }
    cout << foo;
}

**VirtualAce** · 06-06-2010

The problem arises when the name is a macro name, since macros do not obey the rules of scope. Otherwise, one can have a name in the global namespace be the same as a name in some other namespace.

Very true. My example was a poor one.

How about this one then:

Code:

class A 
{
    public:
      A();
      virtual ~A();
   
   private:
     int m_myVariable;
};

class B : public A
{
    public:
      B();
      virtual ~B();

    private:
      int m_myVariable;
};

**laserlight** · 06-07-2010

Originally Posted by Bubba

Very true. My example was a poor one.

How about this one then:

As an example of name hiding, that's fine, though I like the example where the name of an overloaded member function hides the names of member function inherited from the base class. But if you mean it as an example of where namespaces do not help, then that is true, but rather a moot point since m_myVariable is private. If it were non-private, then one could qualify the name and so access A::m_myVariable, and thus effectively get the same benefit as the use of a namespace.

**shiroaisu** · 06-07-2010

Ok, tks everyone i now understand

and ty for the example iMalc

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