Copy a pointer to an element of an array of pointers

[Majid Shahabfar]

How can I copy a pointer to an item of an array pointer such as below? ClassA *ptr = new ClassA[5]; ClassA *ptr1 = somePointerOFClassA; ClassA *ptr2 = somePointerOFClassA; ptr[0] = ptr1; // Error &(ptr[1]) = ptr2; // Error What should I do?

[Hans Ruck]

The array pointer declaration could be:

ClassA **ptr = new ClassA*[5];

Then use ptr[i] = ClassA_Ptr_. rechi

[David Crow]

ptr[0] is an actual object, not a pointer to an object. Therefore, it needs to be assigned the value of ptr1, not the address of ptr1. ptr[0] = *ptr1;

[jhaga]

If you want an array of pointers then do like this ClassA *ptr[5]; ClassA *ptr1 = somePointerOFClassA; ClassA *ptr2 = somePointerOFClassA; and now ptr[0]=ptr1; jhaga

[Peter Mares]

the line ClassA *ptr = new ClassA[5]; defines an array of 5 ClassA objects, not an array of 5 ClassA pointers. If you want an array of 5 ClassA pointers, your declaration will have to be: ClassA **ptr = new ClassA*[5]; and then you will be able to execute the following code (or similar): ClassA *ptr1 = somePointerOFClassA; ClassA *ptr2 = somePointerOFClassA; ptr[0] = ptr1; ptr[1] = ptr2; Now, coming back to your previously mentioned code block.Unfortunately, once you have defined an array as you have : ClassA *ptr = new ClassA[5]; You cannot actually change the address of an element within that array, ie, you cannot do this: &ptr[0] = ptr1

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Hope this helps some. Cheers, Peter ------------------------------------------- 99 little bugs in the code, 99 little bugs, Fix 1 bug, recompile.... 101 little bugs in the code...