Why does a function double dereference arguments stored on stack and how is that possible? [duplicate] - windows

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Closed 7 months ago.
I tried to understand "lfunction" stack arguments loading to "flist" in following assembly code I found on a book (The book doesn't explain it. Code compiles and run without errors giving intended output displaying "The string is: ABCDEFGHIJ".) but I can't grasp the legality or logic of the code. What I don't understand is listed below.
In lfunction:
Non-volatile (as per Microsoft x64 calling convention) register RBX is not backed up before 'XOR'ing. (But it is not what bugs me most.)
In portion ";arguments on stack"
mov rax, qword [rbp+8+8+32]
mov bl,[rax]
Here [rbp+8+8+32] dereferences corresponding address stored in stack so RAX should
be loaded with value represented by'fourth' which is char 'D'(0x44) as per my understanding (Why qword?). And if so, what dereferencing char 'D' in second line can possibly mean (There should be a memory address to dereference but 'D' is a char.)?
Original code is listed below:
%include "io64.inc"
; stack.asm
extern printf
section .data
first db "A"
second db "B"
third db "C"
fourth db "D"
fifth db "E"
sixth db "F"
seventh db "G"
eighth db "H"
ninth db "I"
tenth db "J"
fmt db "The string is: %s",10,0
section .bss
flist resb 14 ;length of string plus end 0
section .text
global main
main:
push rbp
mov rbp,rsp
sub rsp, 8
mov rcx, flist
mov rdx, first
mov r8, second
mov r9, third
push tenth ; now start pushing in
push ninth ; reverse order
push eighth
push seventh
push sixth
push fifth
push fourth
sub rsp,32 ; shadow
call lfunc
add rsp,32+8
; print the result
mov rcx, fmt
mov rdx, flist
sub rsp,32+8
call printf
add rsp,32+8
leave
ret
;––––––––––––––––––––––––-
lfunc:
push rbp
mov rbp,rsp
xor rax,rax ;clear rax (especially higher bits)
;arguments in registers
mov al,byte[rdx] ; move content argument to al
mov [rcx], al ; store al to memory(resrved at section .bss)
mov al, byte[r8]
mov [rcx+1], al
mov al, byte[r9]
mov [rcx+2], al
;arguments on stack
xor rbx,rbx
mov rax, qword [rbp+8+8+32] ; rsp + rbp + return address + shadow
mov bl,[rax]
mov [rcx+3], bl
mov rax, qword [rbp+48+8]
mov bl,[rax]
mov [rcx+4], bl
mov rax, qword [rbp+48+16]
mov bl,[rax]
mov [rcx+5], bl
mov rax, qword [rbp+48+24]
mov bl,[rax]
mov [rcx+6], bl
mov rax, qword [rbp+48+32]
mov bl,[rax]
mov [rcx+7], bl
mov rax, qword [rbp+48+40]
mov bl,[rax]
mov [rcx+8], bl
mov rax, qword [rbp+48+48]
mov bl,[rax]
mov [rcx+9], bl
mov bl,0 ; terminating zero
mov [rcx+10], bl
leave
ret
Additional info:
I cannot look at register values just after line 50 which
corresponds to "XOR RAX, RAX" in lfunc because debugger auto skips
single stepping to line 37 of main function which corresponds to
"add RSP, 32+8". Even If I marked breakpoints in between
aforementioned lines in lfunc code the debugger simply hangs so I
have to manually abort debugging.
In portion ";arguments on stack"
mov rax, qword [rbp+8+8+32]
mov bl,[rax]
I am mentioning this again to be more precise of what am asking because question was marked as duplicate and
provided links with answers that doesn't address my specific issue. At line
[rbp+8+8+32] == 0x44 because clearly, mov with square brackets dereferences reference address (which I assume 64bit width) rbp+3h. So, the size of 0x44 is byte. That is why ask "Why qword?" because it implies "lea [rbp+8+8+32]" which is a qword reference, not mov. So if [rbp+8+8+32] equals 0x44, then [rax] == [0x0000000000000044], which a garbage ( not relevant to our code here) address.