Assembly x86 brk() call use

Question

I am trying to dynamically allocate memory into the heap and then assign values in those memory addresses. I understand how to allocate the memory but how would I assign for example the value in a register to that first dynamic memory address? This is what I have so far:

    push rbp
    mov rbp, rsp            ;initialize an empy stack to create activation records for the rest of the subroutines                                                                                                                        

    mov rax, 0x2d           ;linux system call for brk()                                                                                                                                                                                  
    mov rbx, 0x0            ;to get the adress of the first adress we are allocating we must have 0 in rbx                                                                                                                                
    int 0x80                ;calls the linux operating system kernel for assistance                                                                                                                                                       
    mov [brk_firstLocation], rax ;the first position in the heap will be returned in rax thus i save the first loaction in a varable called brk_firstLocation                                                                             

    mov rbx, rax            ;the memory adress of the start of the heap is moved in rbx                                                                                                                                                   
    add rbx, 0x14           ;we want 5 bytes worth of data alocated in the heap, so the start adress plus 20 bits                                                                                                                         
    mov rax, 0x2d           ;linux system call for brk()                                                                                                                                                                                  
    int 0x80                ;calls the linux operating system kernel for assistance

What would I do, for example, to mov the value in rax into brk_firstLocation

Answer 1

others have pointed out a few things that are wrong with your code. I would like to add that you would not add 20 bits to the current breakpoint (or 20 bytes like add rbx, 20 actually does), you would simply add 5 bytes.

Also, your first syscall argument will not be in rbx, it will be in rdi. The 64-bit syscall ABI uses different system call numbers, different registers, and a different instruction (syscall instead of int 0x80) than the 32-bit ABI (which is still available in 64-bit processes). See also the x86 tag wiki for more ABI links.

Here's how your code should look:

push rbp
mov rbp, rsp

;; sys_brk(0)
mov   rax, 12         ; 12 is SYS_brk (/usr/include/asm/unistd_64.h)
mov   rdi, 0          ; rdi for first syscall arg in the 64-bit ABI, not rbx
syscall               ; syscall, not int 0x80, for the 64-bit ABI

mov   qword [brk_firstLocation], rax

;; sys_brk(old_break + 5)
lea   rdi, [rax + 5]  ; add 5 bytes to the break point
mov   rax, 12
syscall               ; set the new breakpoint

---

At this point you can use brk_firstLocation as a pointer to whatever 5 byte struct you want to store on the heap. Here's how you would put values in that memory space:

mov   rdi, [brk_firstLocation]   ; load the pointer from memory, if you didn't already have it in a register

mov   byte [rdi], 'A'            ; a char in the first byte
mov   [rdi+1], ecx               ; a 32-bit value in the last 4 bytes.

Answer 2

int 80h is only for 32 bit system calls. Use syscall for 64 bit instead.

Calling sys_brk twice is redundant - in assembly you always know where your program data ends. Simply put a label there and you will have the address.

Allocating this way memory less than one page is pointless, it will be allocated in blocks of 4KB anyway.

It is important to be understood - sys_brk is not heap management function. It is low level memory management.

Answer 3

There are a number of problems I see:

• 0x2d is the brk system call on x86 (32 bit); on x86_64 it's 0xc
• brk sets the end of the data segment; it returns 0 on success and -1 on failure. It does not return "the first position in the heap". That comes from the symbol _end which the linker sets to the end of the uninitialized preallocated data.

So you want something like:

    mov [brk_firstloaction], _end
    mov rbx, [brk_firstlocation]
    add rbx, 0x14         ; space for 5 dwords (20 bytes)
    mov rax, 12
    int 0x80