Home >Backend Development >C++ >Why Do Multidimensional Arrays Decay to Pointers Differently Than Single-Dimensional Arrays?
When working with arrays and pointers, it's important to understand how type-decay occurs. While you may expect two-dimensional arrays to decay into double pointers, this isn't always the case. Let's delve into why this happens and explore the difference in behavior.
As the test case demonstrates, one-dimensional arrays indeed decay into single pointers:
<code class="cpp">std::is_same<int*, std::decay<int[]>::type>::value; // true</code>
This is because pointer arithmetic can be performed with a single pointer.
However, two-dimensional arrays don't decay into double pointers:
<code class="cpp">std::is_same<int**, std::decay<int[][1]>::type>::value; // false</code>
The reason is that double pointers require additional information about the dimensions of the array. For instance, in the case of int[5][4], the compiler knows that each "inner" array has a length of 4. Casting to int (*)[4] retains this information, making pointer arithmetic possible.
However, casting to int ** loses this dimension information. It becomes simply a pointer to a pointer, which isn't enough to perform meaningful pointer arithmetic.
Consider the following:
<code class="cpp">char *tmp = (char *)p // Work in units of bytes (char) + i * sizeof(int[4]) // Offset for outer dimension (int[4] is a type) + j * sizeof(int); // Offset for inner dimension int a = *(int *)tmp; // Back to the contained type, and dereference</code>
This code manually performs array access, demonstrating that the compiler relies on dimension information. int** doesn't provide this information, making it unsuitable for pointer arithmetic.
While one-dimensional arrays decay into single pointers, multi-dimensional arrays do not decay into double pointers because they lack the necessary dimension information. This behavior ensures that meaningful pointer arithmetic remains possible with single-dimension pointers.
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