Passing a pointer to a function

This tip submitted by Irfaan Kamo on 2012-05-20 18:27:57. It has been viewed 32082 times.
Rating of 5.4 with 151 votes

void function(int* p)
//Changing the address to where p points.
int* number = new int(5);
p = number;

int main()
int* pointer = new int(1);
cout << *pointer;

When we run the program it will print out 1. This is because the function creates a local copy of the pointer, so whatever it does within the function doesnt affect the real pointer.
In order to get around this we can pass the pointer by reference:

void function(int* &p)

This will ensure that we work with the actual pointer being passed and not just a copy of it.

(Note that its always a good idea to delete the pointer before reassigning it to another address since not doing so will result in memory leaks.)

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