OS X - x64: stack not 16 byte aligned error

Question

I know that OS X is 16 byte stack align, but I don't really understand why it is causing an error here.

All I am doing here is to pass an object size (which is 24) to %rdi, and call malloc. Does this error mean I have to ask for 32 bytes ?

And the error message is:

libdyld.dylib`stack_not_16_byte_aligned_error: -> 0x7fffc12da2fa <+0>: movdqa %xmm0, (%rsp) 0x7fffc12da2ff <+5>: int3

libdyld.dylib`_dyld_func_lookup: 0x7fffc12da300 <+0>: pushq %rbp 0x7fffc12da301 <+1>: movq %rsp, %rbp

Here is the code:

Object_copy:
    pushq %rbp
    movq %rbp, %rsp

    subq $8, %rsp
    movq %rdi, 8(%rsp)          # save self address
    movq obj_size(%rdi), %rax   # get object size
    imul $8, %rax          
    movq %rax, %rdi 
    callq _malloc             <------------------- error in this call

    # rsi old object address
    # rax new object address
    # rdi object size, mutiple of 8

    # rcx temp reg

    # copy object tag
    movq 0(%rsi), %rcx
    movq %rcx, 0(%rax)

    # set rdx to counter, starting from 8
    movq $8, %rdx

    # add 8 to object size, since we are starting from 8
    addq $8, %rdi

    start_loop:
        cmpq %rdx, %rdi
        jle end_loop

        movq (%rdx, %rsi, 1), %rcx
        movq %rcx, (%rdx, %rax, 1)

        addq $8, %rdx
        jmp start_loop

    end_loop:
        leave 
        ret



Main_protoObj:
    .quad    5                          ; object tag
    .quad    3                          ; object size
    .quad    Main_dispatch_table        ; dispatch table

_main:
    leaq Main_protoObj(%rip), %rdi
    callq Object_copy                # copy main proto object
    subq $8, %rsp                    # save the main object on the stack
    movq %rax, 8(%rsp)
    movq %rax, %rdi                 # set rdi point to SELF
    callq Main_init
    callq Main_main

    addq $8, %rsp                    # restore stack

    leaq _term_msg(%rip), %rax
    callq _print_string

Answer 1

Like you said, MacOS X has a 16 byte stack alignment, which means that the machine expects each variable on the stack to start on a byte that is a multiple of 16 from the current stack pointer.

When the stack is misaligned, it means we start trying to read variables from the middle of that 16 byte window and usually end up with a segmentation fault.

Before you call a routine in your code, you need to make sure that your stack is aligned correctly; in this case, meaning that the base pointer register is divisible by 16.

subq $8, %rsp               # stack is misaligned by 8 bytes
movq %rdi, 8(%rsp)          #
movq obj_size(%rdi), %rax   #
imul $8, %rax               #
movq %rax, %rdi             #
callq _malloc               # stack is still misaligned when this is called

To fix this, you can subq the %rsp by something like 16 instead of 8.

subq $16, %rsp               # stack is still aligned
movq %rdi, 16(%rsp)          #
...                          #
callq _malloc                # stack is still aligned when this is called, good