char* and read-only memory

Question

Take a function that creates a char* and does something with it.

void print_hello_twice()
{
    char* hello = "hello";
    printf(hello);
    printf(hello);
}

This is a silly pointles example, but I wanted to avoid anything extraneous to my question, which is this:

I've read that when you assign a string literal to char* variable:

The actual string literal "hello" is stored in the read only memory section. The variable "word" though, is just a pointer created in the stack that as a value, holds the address of the first element of the string.

Memory allocation when using struct to defined a user defined type

So, does this mean that this is a potential source of a memory leak? What happens to the piece of memory containing "hello" itself (not the pointer to it) once print_hello_twice returns? Is it freed up in some way?

Answer 1

Yes and no. Yes it is stored to memory, but the way in which it is declared has no memory leaks (try making a test file and running valgrind to see for yourself).

What does cause a memory leak is when you allocate memory for a variable using malloc (or similar) and don't subsequently free that memory. For example:

void print_hello_twice()
{
    char* hello = malloc(6);
    hello = "hello";
    printf(hello);
    printf(hello);

    // you must free alloc'ed memory when finished
    free(hello);
}

Now doing it this way in such a small program seems a little unnecessary, but for much larger and more complex programs it is, in fact, essential.