Question

This is a question that came up in an exam paper I’m studying for.

“Study the code below. Draw a diagram depicting the organisation of a 32-bit
Linux process address space. In your diagram label the approximate locations
of: a, b, c, *c (i.e. the memory pointed to by c), malloc and main”

int a; int b = 10;
int main()
{
  char *c;
  c = malloc(b);
  return (0);
}

Here’s my take on it, please correct me if I’m wrong.

a is an uninitialized global variable and so would be initialised to zero and stored in BSS. b is an initialized global variable and so would go on the data segment. *c is an automatic variable, so would be stored on the stack. c is dynamically allocated, so would go on the heap.
Main is execution code, so would be stored in text.

I’m not sure on the malloc part, is it just program code that would go in text also, or would it be on the stack as a function call?

Answer 1

You are mostly right about the variables. Just one small correction though. c refers to the pointer variable and *c refers to the memory pointed to by c. So c is an automatic variable and goes on stack and *c is dynamically allocated memory so it resides in heap.

About malloc, the function call maybe made with the help of stack i.e the argument you are passing to malloc may be passed using stack but the function i.e the executable part, it resides in text section. What would happen is that main() would create argument for malloc() on stack, set its value and then set PC=address of malloc(). It is also possible that the argument to malloc is passed using a register instead of stack as there is only one argument but that may vary from compiler to compiler and platform to platform.