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Problems handing a pointer as a variable

Started by Rasmadrak Sep 25, 2007 at 2:26 AM 4 replies 1.2k views

Original Post

Sep 25, 2007 at 2:26 AM

Hi, I'm curious, why doesnt' the following function work? or rather, why doesn't the allocated memory stay beyond the scope of the function that created it?




void doStuff(float *Kalle)
{
   Kalle = new float[100];
   //Kalle == float[100]...
}

//...

main loop:
{
   doStuff(Kalle);
   //Kalle == 0x0.... wtf?
}

I can easily bypass this problem by creating float[100] and handing that to the doStuff function, but that doesn't allow for a clean user-transparent code. So, anyone? /Robert

ToohrVyk

Sep 25, 2007 at 2:33 AM

C++ passes arguments by value, unless you specify otherwise. In your case, passing your pointer by reference (char* &Kalle) would work, though passing a reference to a vector would solve the ownership ambiguities as well.

instinKt

Sep 25, 2007 at 2:37 AM

When you pass your pointer to the function, the pointer is _copied_. Once in the function you are no longer manipulating the original pointer but a copy of it. Thus, when you return, that copy is destroyed (your allocated memory will be lost causing a memory leak) and you're left with the original pointer.

To do what you're wanting, you need to pass a pointer to the pointer.

void doStuff(float **Kalle){   *Kalle = new float[100];   //Kalle == float[100]...}//...main loop:{   doStuff(&Kalle);   //Kalle == 0x0.... wtf?}

Notice the double ** (pointer to pointer), when new is called you're dereferencing the first pointer with the single *, and when you pass the pointer to the function you give it the address of the original pointer using &.

It can get a bit confusing, but the more you play around with it, the more you'll understand.

Daishim

Sep 25, 2007 at 2:38 AM

The allocated memory actually does exist beyond the doStuff function, however, a little pointer explanation is in order. What you are passing to doStuff is a memory address, which it copies to your pointer that becomes the argument. The argument variable is then assigned new memory, and then points to a new memory location, while your original variable is untouched. To correct this, you want to pass a pointer reference, like so:

void doStuff(float *&Kalle){     Kalle = new float[100];}int main(void){     float *My_Stuff = NULL;     doStuff(My_Stuff);     ...}

Rasmadrak

Sep 25, 2007 at 2:41 AM

Hi!

Thanks for your replies, it all makes sense now - actually I'm rather embarresed that it was that simple. :)

Pointers - can't live with them, can't kill them!

Thanks again!

/Robert

instinKt

Sep 25, 2007 at 2:43 AM

Quote:
Original post by Rasmadrak
Pointers - can't live with them, can't kill them!

Sure you can! It just hurts you more than them. :(

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