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What does a reference to a pointer store?

waqqassheikh (100)
I'm a little confused about what a pointer to a reference is. How would you describe it. For example:

[code
int n;
int* m = &n[/code]
Here pointer m stores the memory address of an integer 'n'.

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int n;
int& m = n

Here m stores the memory address of integer 'n'.

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int n;
int* j = &n
int*& m = j

j is a pointer which stores the memory address of integer 'n'.
What is m in this case? What does it store?

Second, is there a difference between passing a pointer: 
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int main()
{
int n = 5;
int* m = &n;
foo(m);
}

void foo(int* j)
{
//
}


...and passing a pointer to a reference?
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int main()
{
int n = 5;
int* m = &n;
foo(m);
}

void foo(int*& j)
{
//
}


Thanks Guys!
Last edited on
Disch (13742)
Conceptually, a reference "is" the variable it refers to.

Therefore

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int n;
int& m = n;


Here, m "is" n. They are just different names for the same variable.

With that in mind:

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int n;
int* j = &n;
int*& m = j;


Here, j points to n
And m "is" j. Thererfore m also points to n, because m and j are one and the same.


EDIT:

Also int*& is a reference to a pointer, not a pointer to a reference. ;P
Last edited on
waqqassheikh (100)

EDIT:

Also int*& is a reference to a pointer, not a pointer to a reference. ;P

Thanks for making that clear ... and for your answer - but I'm still a little confused:

If m and j are one and the same, then why are there situations where one can't be used instead of another. For instance, in this code

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#include <iostream>
using namespace std;

void print(char* );
void replaceWordWith(char*&, char*);

int main()
{
char* s = "Hello";
print(s);
replaceWordWith(s,"Bye");
print(s);
return 0;
}

void print(char* c)
{
cout<<c<<endl;
}

void replaceWordWith(char*& c, char* s)
{
c = s;
}


Output

Hello
Bye


I don't think a simple pointer could replace the char*& c here
Last edited on
Disch (13742)
I don't think a simple pointer could replace the char*& c here


No no... pointers and references are not one and the same. The reference is one and the same with the object it refers to.

In your example....

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void replaceWordWith(char*& c, char* s);

replaceWordWith(s,"Bye");


Here, since replaceWordWith's 'c' parameter is a reference, c "is" whatever gets passed to it. Since you're passing it 's', this means that in replaceWordWith, c "is" s. This is why when you change c, s is changed. Because they're just different names for the same variable.

You are correct in thinking that if you replace c with a regular pointer (and not a reference to a pointer) then the same thing will not be accomplished. That is because doing so will make c a separate variable from s, and changing c will not change s.
waqqassheikh (100)
That's an excellent explanation!
Thanks, really appreciate your help.
I think I've got it now :)
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