I lose pointer or an address? It acts like never happen

Question

So the problem is I have a function for example

int * fun2(){...}

int fun(int *a)
{
   a = fun2(); // this is like line never exist
}

I want to assign returned pointer by fun2 to pointer a and it doesnt happen WHY?

And an "a" variable has already a value, before assigning fun2(), but I dont think so it really matter.

If you are asking "why does `a` still have the old value after I call `fun(a)`, see [this question](http://stackoverflow.com/questions/4426474/is-passing-pointer-argument-pass-by-value-in-c). — DCoder, Dec 29 '13 at 19:32

score 1 · Answer 1 · answered Dec 29 '13 at 19:32

It does work. It changes the value of a in the local scope.

If you want to see change outside of fun, you have to pass a reference: int fun(int *&a).

(Note: you have to provide us a complete example which demonstrates your problem, so we don't have to do educated guesses).

score 1 · Accepted Answer · answered Dec 29 '13 at 19:32

1

Your a is a temporary object in fun(). I fyou want fun() to change the pointer to a you'll need to take the pointer as reference:

int fun(int*& a) {
    a = fun2();
}

Note that this interface is probably severely broken.

answered Dec 29 '13 at 19:32

Dietmar Kühl

score 0 · Answer 3 · answered Dec 29 '13 at 20:18

0

Another alternative is to pass the address of the a pointer variable:

int fun(int **a)
{
  *a = func2();
}

answered Dec 29 '13 at 20:18

Thomas Matthews

score 0 · Answer 4 · answered Dec 29 '13 at 23:52

0

You passed the pointer by value. Inside the function, it's a new copy of that pointer. You change only the copy in there.

answered Dec 29 '13 at 23:52

4 Answers4