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

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

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