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Why Does Array Decay Differ for int[] and int[][] in C ?

Mary-Kate Olsen
Mary-Kate OlsenOriginal
2024-10-26 04:15:02571browse

 Why Does Array Decay Differ for int[] and int[][] in C  ?

Type Decay in Arrays: int* vs. int[][]

In C , arrays decay into pointers in certain contexts, such as when passing them to functions or assigning them to pointers. However, this decay behavior can vary depending on the dimensionality of the array.

Consider the following code:

<code class="cpp">std::is_same<int*, std::decay<int[]>::type>::value; // true
std::is_same<int**, std::decay<int[][1]>::type>::value; // false</code>

The first line compiles successfully, confirming that an int[] array decays into an int pointer. However, the second line fails, indicating that an int[][1] two-dimensional array does not decay into an int* pointer.

The reason for this discrepancy lies in the nature of pointer arithmetic. In a single-dimensional array like int[], each element is stored consecutively in memory. This allows a pointer to the array to be incremented or decremented to access adjacent elements. However, in a two-dimensional array like int[][1], each element is stored as a pointer to a subarray of ints. If an int** pointed to this two-dimensional array, it would not be able to perform meaningful pointer arithmetic because it would not know the size of each subarray.

Instead, to maintain the relationship between the dimensions of an array when it decays, C decays a two-dimensional array into a pointer to an array of pointers. In other words, int[][1] decays into int (*[])[1].

This behavior applies to arrays of any dimensionality, including mixed arrays with pointers and non-pointers. For example, int[] decays into int but not int[][], while int[] decays into int.

Understanding this type decay mechanism is crucial when working with arrays in C to ensure compatibility between array types and pointer types.

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