pass a dynamic array to a function 1

[stranger123]

Hi,

I remember a book said we can use *& to pass an array to a function, but now I find it can be done by use * only. For example:

void main()
{
int* myArray = new int[1];
myArray[0]=1;
MyFunction(myArray);
cout<<myArray[0]<<endl;
delete[] myArray;
}
void MyFunction(int* myArray) //void MyFunction(int*& myArray)
{
myArray[0]++;
}

So, what is the difference between * and *& when pass an array to a function?

Can you help?

 

[cpjust]

* passes a copy of a pointer.
*& passes a reference to a pointer.

Basically if MyFunction() assigned a different pointer value like this:

void MyFunction( int* myArray )
{
   delete [] myArray;
   myArray = new int[ 50 ];
}

The myArray in your main() function would still point to the same address that it did before calling MyFunction() (which is now invalid because it points to memory that was deleted).

If you changed it to *& then myArray in main() would point to the new address that was allocated by MyFunction().

 

[stranger123]

Thanks a lot for your help!