Stack memory fundamentals

Question

Consider this code:

 char* foo(int myNum) {
    char* StrArray[5] = {"TEST","ABC","XYZ","AA","BB"};

    return StrArray[4];
 }

When I return StrArray[4] to the caller, is this supposed to work? Since the array is defined on the stack, when the caller gets the pointer, that part of memory has gone out of scope. Or will this code work?

Answer 1

This code will work. You are returning the value of the pointer in StrArray[4], which points to a constant string "BB". Constant strings have a lifetime equal to that of your entire program.

The important thing is the lifetime of what the pointer points to, not where the pointer is stored. For example, the following similar code would not work:

char* foo(int myNum) {
   char bb[3] = "BB";
   char* StrArray[5] = {"TEST","ABC","XYZ","AA",bb};

   return StrArray[4];
}

This is because the bb array is a temporary value on the stack of the foo() function, and disappears when you return.

Answer 2

Beware: you're lying to the compiler.

Each element of StrArray points to a read-only char *;
You're telling the compiler the return value of your function is a modifiable char *.
Lie to the compiler and it will get its revenge sooner or later.

In C, the way to declare a pointer to read-only data is to qualify it with const.

I'd write your code as:

const char* foo(int myNum) {
   const char* StrArray[5] = {"TEST","ABC","XYZ","AA","BB"};

   return StrArray[4];
}