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Why Does Type Decay Differ for One-Dimensional and Multidimensional Arrays in C ?

Susan Sarandon
Release: 2024-10-26 11:48:29
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Why Does Type Decay Differ for One-Dimensional and Multidimensional Arrays in C  ?

Type Decay in Multidimensional Arrays

In C , arrays undergo type decay into pointers in certain contexts. However, the behavior differs for one-dimensional arrays and multidimensional arrays. Why is this the case?

One-Dimensional Arrays

Consider the following code:

<code class="cpp">int[] arr = {1, 2, 3};</code>
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Type decay converts int[] to int*, allowing us to write:

<code class="cpp">std::is_same<int*, std::decay<int[]>::type>::value; // true</code>
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This is because int[] effectively decays into a pointer to the first element of the array.

Multidimensional Arrays

However, this behavior does not extend to multidimensional arrays. Consider:

<code class="cpp">int[][1] arr2 = {{1}, {2}, {3}};</code>
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Type decay does not convert int[][1] into int**. Instead, it remains int[][1], which is a pointer to an array of size 1. This is because performing pointer arithmetic on a multidimensional array would be impractical due to its non-contiguous memory layout.

Pointers to Arrays

In order to obtain the desired pointer type for multidimensional arrays, we need to create pointers to arrays. Consider:

<code class="cpp">int*[] arr3 = {arr, arr2};</code>
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This decays into:

<code class="cpp">int**</code>
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This is because int*[] decays into int**, as the first dimension is a pointer and the second dimension is an array.

Implications

Understanding this behavior is crucial for passing multidimensional arrays as function arguments. Functions expecting pointers to pointers cannot accept multidimensional arrays themselves, but they can accept pointers to arrays, which decay into pointers to pointers during parameter passing. This subtle distinction ensures that pointer arithmetic is consistent and prevents memory access errors.

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