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[Assembly] asm source code of the program written in C? - Printable Version

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+---- Thread: [Assembly] asm source code of the program written in C? (/showthread.php?tid=5861680)

asm source code of the program written in C? - SecuresSpy - Jul 14, 2018

What is the assembly source code of the following C code?

char string[1001];

while(string[index] != '\0') {
diff = string[index] - string[index - 1];
if(diff < 0) {
count += (26 + diff);
} else {
count += diff;
}
index += 1;
}

hints:
PHP Code

<?php 
loc_4013B4:
1. lea     edx, [esp+13h]
2. mov     eax, [esp+408h]
3. add     eax, edx
4. movzx   eax, byte ptr [eax]
5. movsx   edx, al
6. mov     eax, [esp+408h]
7. sub     eax, 1
8. movzx   eax, byte ptr [esp+eax+13h]
9. movsx   eax, al
sub     edx, eax
mov     eax, edx
mov     [esp+400h], eax
cmp     dword ptr [esp+400h], 0
jns     short loc_401401 

note:
line 1 loading address of array from stack.
what the 2 and 3 and others line are doing?

Thanks

RE: asm source code of the program written in C? - Vanilla - Jul 27, 2018

(Jul 14, 2018 06:29 AM)SecuresSpy Wrote:  What is the assembly source code of the following C code?

char string[1001];

   while(string[index] != '\0') {
    diff = string[index] - string[index - 1];
    if(diff < 0) {
    count += (26 + diff);
    } else {
    count += diff;
    }
    index += 1;
    }

hints:
PHP Code

<?php 
loc_4013B4:
1. lea     edx, [esp+13h]
2. mov     eax, [esp+408h]
3. add     eax, edx
4. movzx   eax, byte ptr [eax]
5. movsx   edx, al
6. mov     eax, [esp+408h]
7. sub     eax, 1
8. movzx   eax, byte ptr [esp+eax+13h]
9. movsx   eax, al
sub     edx, eax
mov     eax, edx
mov     [esp+400h], eax
cmp     dword ptr [esp+400h], 0
jns     short loc_401401 

note:
line 1 loading address of array from stack.
what the 2 and 3 and others line are doing?

Thanks

Ok, your code was completely screwed up but added some stuff to make it compile.

C code:
Code
char string[1001];

int main() {
    int index;
    int diff;
    int count;
    while(string[index] != '\0') {
        diff = string[index] - string[index - 1];
        if (diff < 0) {
            count += (26 + diff);
        } else {
            count += diff;
        }
    }
    index += 1;
}

ASM code:
Code
main:
        push    rbp
        mov     rbp, rsp
        jmp     .L2
.L4:
        mov     eax, DWORD PTR [rbp-8]
        cdqe
        movzx   eax, BYTE PTR string[rax]
        movsx   edx, al
        mov     eax, DWORD PTR [rbp-8]
        sub     eax, 1
        cdqe
        movzx   eax, BYTE PTR string[rax]
        movsx   eax, al
        sub     edx, eax
        mov     eax, edx
        mov     DWORD PTR [rbp-12], eax
        cmp     DWORD PTR [rbp-12], 0
        jns     .L3
        mov     eax, DWORD PTR [rbp-12]
        add     eax, 26
        add     DWORD PTR [rbp-4], eax
        jmp     .L2
.L3:
        mov     eax, DWORD PTR [rbp-12]
        add     DWORD PTR [rbp-4], eax
.L2:
        mov     eax, DWORD PTR [rbp-8]
        cdqe
        movzx   eax, BYTE PTR string[rax]
        test    al, al
        jne     .L4
        add     DWORD PTR [rbp-8], 1
        mov     eax, 0
        pop     rbp
        ret

Enjoy!

RE: asm source code of the program written in C? - DebugBSD - Sep 8, 2018

Hi, good afternoon!

This is another version of the same algorithm. I've tested with random data and it seems to work propertly.

Code
.data

string BYTE 1001 DUP(0)
len EQU($ - string)

.code
main PROC

   mov RDI, OFFSET string
   mov RCX, 1001
   ; Load random data
L2:
   RDRAND RAX
   stosb
   loop L2

   mov RDI, OFFSET string
   mov RDX, 0
L1:
   scasb
   je EXIT
   mov RBX, [RDI+1]
   sub BL, BYTE PTR [RDI]
   AND RBX, 0FFh
   cmp BL, 0
   jae I1
   add BL, 26
   add RDX, RBX
I1:
   add RDX, RBX
   jmp L1

EXIT:
   call ExitProcess
main ENDP

Enjoy Assembly Language Programming!

RE: asm source code of the program written in C? - Infamia - Sep 24, 2018

I'll go off Hysteric's diassembly:

Code
main:
        push    rbp
        mov     rbp, rsp
        jmp     .L2

Set up an x64 stack-frame, jump to the label .L2:

Code
.L2:
        mov     eax, DWORD PTR [rbp-8]
        cdqe
        movzx   eax, BYTE PTR string[rax]
        test    al, al
        jne     .L4
        add     DWORD PTR [rbp-8], 1
        mov     eax, 0
        pop     rbp
        ret

Code
mov     eax, DWORD PTR [rbp-8]
cdqe

is kinda odd because it'd be better in the form

Code
movq -8(%rbp), %rax
; mov rax, QWORD PTR[rbp-8]

but it's probably some intrinsic detail; regardless it just moves a locally allocated pointer to the eax/rax register

Code
.L2:
movzx   eax, BYTE PTR string[rax]
test    al, al
jne     .L4

add     DWORD PTR [rbp-8], 1
mov     eax, 0
pop     rbp
ret

moves (with zero-extension, i.e the high 4 bytes of %rax are zeroed) the rax'th index of the `string` into `eax` as a character pointer (BYTE has the same range as a character, from 0 to 255), if the lower byte of eax (that is, the character that has just been stored) is not zero (not the end of the string, since C strings are null-terminated by a character whose ordinal is 0x00) then jump to the actual loop, else, add 1 to the index and return 0 (eax is the return variable for the stack-frame in this convention).

note: the `add DWORD PTR [rbp-8], 1` is a flaw of the code, there shouldn't be an `index += 1;`, it doesn't make sense and has no purpose.

Code
.L4:
mov     eax, DWORD PTR [rbp-8]
cdqe

movzx   eax, BYTE PTR string[rax]
movsx   edx, al

mov     eax, DWORD PTR [rbp-8]
sub     eax, 1
cdqe

movzx   eax, BYTE PTR string[rax]
movsx   eax, al

sub     edx, eax

mov     eax, edx
mov     DWORD PTR [rbp-12], eax
cmp     DWORD PTR [rbp-12], 0

jns     .L3

mov     eax, DWORD PTR [rbp-12]
add     eax, 26
add     DWORD PTR [rbp-4], eax
jmp     .L2

firstly the `index` is reloaded into `eax` for some reason, even though it was just loaded prior to that; probably due to a lack of optimization, then the character is loaded back into `eax` as last time, `movsx edx, al` moves the character with sign-extension (i.e., if al is negative then the high four bytes of %rax are 0xFFFF else 0x0000), afterwards the index is loaded again into `eax`, subtracted by 1, converted to a quadword/long long, then the character before is loaded into eax, the difference between the two characters is found by `sub edx, eax`, the result of the subtraction is in `eax`, the old character is moved into the new character, then the character is stored in another locally allocated variable at `rbp-12`, thus the character is compared to 0, if the comparison result of `[rbp-12] - 0` is not signed (i.e. the result is non-negative) then jump to .L3.

else if it's negative then move the old character into eax, add 26 to the character, and then add the previous addition to a new variable which i didn't know existed up until now, hereby reference this as the 1st locally stored variable at rbp-4, didn't take into account that they were DWORD sizes opposed to QWORD so I took a multiple of 8.

afterwards unconditionally jump back to the .L2 label which checks for NULLs, if so, quit, else, quits.

EDIT:

On GCC 8.2 x86-64 with parameters `-O3 -mcmodel=large` (we don't want rip relative addressing fuck that shit)

Code
#include <cstddef>
#include <cstdio>

int main(int argc, char **argv) {
    size_t index;
    size_t diff;
    size_t count;

   char* string = argv[1];

    while(string[index]) {
        diff = string[index] - string[index - 1];
        if (diff < 0) {
            count += (26 + diff);
        } else {
            count += diff;
        }
    }
}

disassembles to

Code
main:
        movq    8(%rsi), %rax
        cmpb    $0, (%rax)
        je      .L2
.L3:
        jmp     .L3
.L2:
        xorl    %eax, %eax
        ret

which is just a bit shorter:

first we move the 1st argument from argv[1] into rax, compare the character at rax by NULL, if equal then jump into an infinite loop, else return 0.

points out the flaw in logic in the code Hysteric wrote.

RE: asm source code of the program written in C? - macheti228 - Nov 20, 2018

good information for те men