Understanding C program in Assembly

Question

I'm trying to understand this simple C program:

int square(int num) {
    return num * num; 
}

When it's in Assembly code:

square(int): 
  push rbp ;push rbp register onto stack
  mov rbp, rsp ;move contents of rbp register into rsp register 
  mov DWORD PTR [rbp-4], edi ;not sure what happens here 
  mov eax, DWORD PTR [rbp-4] ;not sure what happens here
  imul eax, DWORD PTR [rbp-4] ;multiply eax and DWORD PTR [rbp-4] (?)
  pop rbp ;pop original register out of stack
  ret ;return 

1. What is happening in the 3rd and 4th line?
2. Why did two more register (edi and eax) have to be used instead of rsp?
3. What is actually happening with DWORD PTR [rbp-4]?

Answer 1

mov DWORD PTR [rbp-4], edi ;not sure what happens here

The x86_64 System V ABI passes function arguments via registers - the first integer argument is passed in the rdi/edi register. So this line copies the argument num to a local (offset -4 bytes from the frame pointer value stored in rbp).

mov eax, DWORD PTR [rbp-4] ;not sure what happens here

This copies the value in the local to the eax register.

imul eax, DWORD PTR [rbp-4] ;multiply eax and DWORD PTR [rbp-4] (?)

And this multiplies the value in eax by the local, and stores the result to eax (which also happens to be the register in which the function return value is stored).

As others pointed out in the comments, compiling with optimization would likely eliminate the local, and write directly from edi to eax.