Pointer to Reference (if I am not wrong)

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5.00/5 (1 vote)

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Hi,

C++

void incr(int* &f)
{
    f++;
}

int main ()
{

    int *i = 0;
    std::cout<<"i ="<<i;
    incr(i);
    std::cout<<"i ="<<i;

  return 0;
}

I am surprised looking at underlined code, int* &f. Can you explain meaning of this line.

Posted 30-Mar-12 22:05pm

Pranit Kothari

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Code-o-mat · Accepted Answer · 2012-03-30T22:33:00

Solution 1

That's not a pointer to a reference, that's a reference to a pointer.

Posted 30-Mar-12 22:33pm

Code-o-mat

v2

Comments

[no name] 31-Mar-12 4:36am

Thx 5!

nv3 · Accepted Answer · 2012-03-30T22:44:00

The other way around: f is a reference to a pointer. Did you run your example? What you certainly saw is that i is incremented by 4 in the call to incr. That's probably the point to be demonstrated here. f is a reference to a pointer to int, so it behaves like a pointer to int. When you increment it, its value will increase by the number of bytes that equals the size of an int on your computer - usually 4 on 32-bit systems.

Only by passing f by reference made it possible that the change to f is visible in main at all. If you pass it by value like in

void incr(int* f)
{
    f++;
}

f would be incremented, but main would not see the change, as it never would get back there.