How C strings are allocated in memory?

Question

Say I have a simple function that returns a C string this way:

const char * getString()
{
  const char * ptr = \"blah blah\";
  return ptr; 
}

Answer 1

In your function, the scope of a[] array is within the function getString2(). its local array variable.

const char *getString2()
{
  const char a[] = "blah blah blah";
  return a;
}  

In above case string "blah blah blah" copies fist into a[] and your are trying to return that array with return a statement, but not constant string.

Where as in first code getString() : ptr = "blah blah"; ptr point to memory that has global scope.

const char * getString()
{
  const char * ptr = "blah blah";
  return ptr; 
}

In this case you returns the address of constant string "blah blah" that is legal to do.

So actually its Scope problem.

it helpful to learn about Memory Layout of C Programs and Variable Scope in C.