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void incr(int* &f)

int main ()

    int *i = 0;
    std::cout<<"i ="<<i;
    std::cout<<"i ="<<i;

  return 0;

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

That's not a pointer to a reference, that's a reference to a pointer.
[no name] 31-Mar-12 4:36am
Thx 5!
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 would be incremented, but main would not see the change, as it never would get back there.
[no name] 31-Mar-12 4:46am
Correct.. 5!

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