Why Does My Merged Array Disappear After Returning from a Function in C ?

Passing Arrays: Potential Pitfalls

While returning arrays from functions in C appears straightforward, hidden complexities can arise, as exemplified by the user's experience trying to merge non-negative numbers from two arrays using a function.

The user observed incorrect output when accessing the merged array outside the function that created it. This is because arrays created on the stack, like the one declared as "int c[10]" within the function, are destroyed once the function exits. Accessing them afterwards results in undefined behavior.

To address this, one option is to replace the local array with a struct that contains the array. When the struct is returned, its internal array is copied into the calling function's memory, effectively avoiding the stack destruction issue. Here's an example:

<code class="cpp">struct myArray {
    int array[10];
};

myArray uni(int *a, int *b) {
    myArray c;
    // Same logic as before for merging arrays
    return c;
}</code>

In main, the returned struct can be assigned and the internal array accessed:

<code class="cpp">myArray c = uni(a, b);
for (int i = 0; i <p>Another solution is to use a pointer to allocate memory for the array dynamically, as discussed in your "passing pointers" chapter. However, this introduces the need to manage memory manually and avoid memory leaks.</p>
<p>The choice of approach depends on specific requirements, but understanding these pitfalls is crucial when working with arrays in functions.</p></code>

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