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 < 10; i++) {
    cout << c.array[i] << " ";
}</code>

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.

The choice of approach depends on specific requirements, but understanding these pitfalls is crucial when working with arrays in functions.

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