Understanding pointers and local scope [duplicate]

Answer 1

char str[20] = "Hello world.";

str is local to function allocateMemory() and is no more valid once you exit the function and hence accessing it out of its scope if undefined behavior.

int x = 5;

The same applies here also.

You can have your data on heap and return the pointer to it is valid.

char *allocatememory()
{
   char *p = malloc(20); /* Now the memory allocated is on heap and it is accessible even after the exit of this function */
   return p; 
}