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Passing a pointer

Hi,

I'm a little confused on what happens when you pass a pointer to some object as a parameter to a function. Something like this:

int main()
{
    MyObj* p = new MyObj;
    Test1(p);
    Test2(&p);
}

void Test1(MyObj* p)
{
    p->SomeFunction();
}

void Test2(MyObj** pp)
{
    (*pp)->SomeFunction();
}


Is there any difference? I think Test1() is getting a copy of the pointer declared in main(), while Test2() is actually getting the same pointer declared in main().

Thanks
0
DJ_AM_Juicebox
Asked:
DJ_AM_Juicebox
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1 Solution
 
Infinity08Commented:
>> I think Test1() is getting a copy of the pointer declared in main()

Correct.


>> Test2() is actually getting the same pointer declared in main().

To be more exact, it gets a pointer to the pointer declared in main(). This means that you can modify the pointer inside the Test2() function, and that modification will be visible in main() :

    void Test2(MyObj** pp)
    {
        delete *pp;
        *pp = new MyObj;
        (*pp)->SomeFunction();
    }

and calling it :

    MyObj* p = new MyObj;
    Test2(&p);
    // p now points to the newly allocated MyObj
0
 
Jaime OlivaresSoftware ArchitectCommented:
the different version have different uses:
in Test1 you can use the pointer to access some data
in Test2 you can use and change the pointer to point to a different data
0
 
peetmCommented:
When a pointer [or anything is - that's not using 'references'], the expression used in the call is evaluated, and the result of that evaluation is passed - so *values*/*copies* of things are passed to called functions.

In your code a copy of value of p is passed to Test1, and a copy of the address of p is passed to Test2.  The 'type' of the first is 'pointer to a MyObj' and a 'pointer to a pointer to a MyObj' respectively.  And, as you don't declared these formal parameters using 'const' you've turned a constant into a variable at each call, i.e., the result of the expression evaluation is a number - a constant.  But that is 'received' in Test1/2 as a variable.  Still, non-matted really, as each variable is initialised with the value obtained in each evaluation.

So, in Text1 p is a copy of the address of a MyObj as a scalar variable, and in Test2 it is a copy of the address of a pointer to a MyObj as a scalar variable.

When Test1/2 return/complete, these local variables [the formal parameters] are destroyed.

Hope that helps.

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Jaime OlivaresSoftware ArchitectCommented:
to extend my last comment:
void Test2(MyObj** pp)
{
    *pp = new MyObj();
}

this allows Test2 to modify the pointer of the calling function.
0
 
Jaime OlivaresSoftware ArchitectCommented:
Sorry, not a realistic example, will generate a memory leak, will be better:

to extend my last comment:
void Test2(MyObj** pp)
{
    delete (*pp);
    *pp = new MyObj();
}
0
 
peetmCommented:
Sans the typos and tidied a little.
=========================

When a pointer [or anything is - that's not using 'references'] is passed, the expression used in the call is evaluated, and the result of that evaluation is what's passed - so *values*/*copies* of things are passed to called functions with using references.

So, in your code, a copy of *value* of p is passed to Test1, and a copy of the [value] *address* of p is passed to Test2.  The 'types' are 'pointer to a MyObj' and a 'pointer to a pointer to a MyObj' respectively.  As you don't declared these formal parameters using 'const' you've turned a constant into a variable at each call, i.e., the result of the expression evaluation is a number - a constant.  But that is 'received' in Test1/2 as a variable.  Still, no matter really, as each local variable is initialised with the value obtained by each of the evaluations taking place ahead of each call.

So, in Test1, p is a copy of the address of a MyObj as a scalar variable, and in Test2 it is a copy of the address of a pointer to a MyObj as a scalar variable.

When Test1/2 return/complete, these local variables [the formal parameters] are destroyed.

Hope that helps.
0
 
Infinity08Commented:
nice example, jaime, but a bit late ;)
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Jaime OlivaresSoftware ArchitectCommented:
Oops, not noticed, I think my proxy is not refreshing correctly. I am arriving late to many questions.
0
 
peetmCommented:
I seem to arrive late at every question usually - what could be the problem?  OR - should I ask this somewhere for some points do you think?
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