Question

I’m running through some assembly code and I can’t figure out what a line of code does. The code is:

 leaq   0(,%rax,4), %rdx

I know lea is basically a type of mov instruction, but it only moves the address. So we are moving the address of something to %rdx (making %rdx “point” to something on the stack). I know what %rax points to on the stack (say, -28(%rbp)), but I’m confused by how to multiply that with 4 to get my answer. Would %rdx point to 4*(-28) = -112(%rbp)?

Thanks!

EDIT:
For context, the following code precedes this instruction:

pushq   %rbp 
movq    %rsp, %rbp 
movl    %esi, -28(%rbp)
movl    -28(%rbp), %eax 
cltq 
leaq    0(,%rax,4), %rdx

Answer 1

Your equivalent C code is something like:

extern int32 arr[];
int my_func(int32 n, ...) {
   int32 a=n;
   ...
   arr[a];
   ...
}

n is passed as a single 32-bit register esi, which is stored to local stack frame.
The parameter is then used in evaluation the 64-bit expression 4*a. The ‘0’ can be explained if it’s supposed to be relocated by the linker to the address ‘arr’.

Then my guess is that the assembly code is not generated by
gcc -S foo.c, but by gcc -c foo.c; objdump -d foo.o

   // Similar code from a 32-bit machine
   0:   55                      push   %ebp
   1:   89 e5                   mov    %esp,%ebp
   3:   8b 45 08                mov    0x8(%ebp),%eax
   6:   8b 04 85 00 00 00 00    mov    0x0(,%eax,4),%eax
   d:   5d                      pop    %ebp
   e:   c3                      ret
   f:   90                      nop