User Defined Array Size

int * aPointer;
aPointer = new int[10][10];

Why won't it let me assign a multi-dimensional array?
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BabyFaceAsked:
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wpintoCommented:
Hi BabyFace,

When new is used to allocate a single object, it yields a pointer to that object; the resultant type is new-type-name * or type-name *. When new is used to allocate a singly dimensioned array of objects, it yields a pointer to the first element of the array, and the resultant type is new-type-name * or type-name *. When new is used to allocate a multidimensional array of objects, it yields a pointer to the first element of the array, and the resultant type preserves the size of all but the leftmost array dimension. For example:

new int[10][10]
yields type int (*)[10].

Therefore, the following code will not work because it attempts to assign a pointer to an array of int with the dimensions [10] to a pointer to type int:

int *aPointer;
aPointer = new int[10][10];

The correct expression is:

int (*aPointer)[10];
aPointer = new int[10][10];

Hope this helps

Wilfred
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ozoCommented:
line 3: error(3611): a value of type "int (*)[10]" cannot be
          assigned to an entity of type "int *"
  aPointer = new int[10][10];
           ^
//try
int (*aPointer)[10];
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BabyFaceAuthor Commented:
Good job.

Thanks.

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