How to Safely Return a Pointer to a Local Variable in C ?

How to Return a Pointer to a Local Variable in C

In C , it is generally discouraged to return a pointer to a local variable, as the variable's memory space may be deallocated when the function exits. However, in certain scenarios, it may be necessary to achieve this.

Consider a scenario where you need to create a function that returns a pointer to an integer:

int *count()
{
    int myInt = 5;

    int * const p = &myInt; // Pointer to the local variable

    return p;
}

In this case, the issue arises because the myInt variable is destroyed when the function exits, leaving the pointer pointing to an invalid memory location.

To address this issue, you can utilize smart pointers, which automatically manage the memory allocated to the variable. For instance, you can use a unique_ptr:

unique_ptr<int> count()
{
    unique_ptr<int> value(new int(5));
    return value;
}

Here, value is a unique_ptr that manages the memory for the dynamically allocated integer. This ensures that the integer's memory will not be deallocated when the function exits, allowing you to access it through the returned pointer:

cout << "Value is " << *count() << endl;

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