ikarus/bin/ikarus-enter.s

126 lines
3.6 KiB
ArmAsm
Raw Normal View History

2006-11-23 19:38:26 -05:00
.text
2006-11-30 18:12:01 -05:00
.globl ik_asm_enter
2006-11-23 19:53:15 -05:00
.globl _ik_asm_enter
2006-11-23 19:38:26 -05:00
.globl ik_underflow_handler
2006-11-30 18:12:01 -05:00
.globl ik_foreign_call
2006-11-23 19:53:15 -05:00
.globl _ik_foreign_call
2006-11-30 18:12:01 -05:00
.globl ik_asm_reenter
2006-11-23 19:53:15 -05:00
.globl _ik_asm_reenter
2006-11-23 19:42:39 -05:00
.align 8
2006-11-30 18:12:01 -05:00
ik_asm_enter:
2006-11-23 19:53:15 -05:00
_ik_asm_enter:
2006-11-23 19:38:26 -05:00
# ignored value is the third arg 12(%esp)
# code is the second arg 8(%esp)
# pcb is the first arg 4(%esp)
# return point is at 0(%esp)
movl %esi, -4(%esp) # preserve
movl %ebp, -8(%esp) # preserve
movl 4(%esp), %esi
movl 0(%esi), %ebp # allocation pointer is at 0(pcb)
movl %esp, %eax
2006-11-23 19:53:15 -05:00
subl $16, %esp # 24 for alignment
set_stack:
2006-11-23 19:38:26 -05:00
movl %esp, 24(%esi) # save esp in pcb->system_stack
movl 8(%esi), %esp # load scheme stack from pcb->frame_pinter
jmp L_call
.byte 0
.byte 0
.byte 0
.byte 0
.byte 0
.byte 0
.byte 0
.byte 0
.long L_multivalue_underflow
.byte 0
.byte 0
L_call:
call *8(%eax) # goooooooo
# now we're back
ik_underflow_handler:
movl %eax, -8(%esp) # store the return value
movl $-4, %eax # set rvcount = 1
L_do_underflow:
movl %esp, 8(%esi) # store scheme stack in pcb->frame_pointer
movl %ebp, 0(%esi) # store allocation pointer
movl 24(%esi), %esp # restore system stack
2006-11-23 19:53:15 -05:00
addl $16, %esp # 24 for alignment (>= 16)
2006-11-23 19:38:26 -05:00
movl -4(%esp), %esi # restore callee-save registers
movl -8(%esp), %ebp #
ret # back to C, which handled the underflow
L_multivalue_underflow:
addl $4, %esp
jmp L_do_underflow
.align 8
2006-11-30 18:12:01 -05:00
ik_asm_reenter:
2006-11-23 19:53:15 -05:00
_ik_asm_reenter:
2006-11-23 19:38:26 -05:00
# argc is at 12(%esp)
# scheme stack is third arg 8(%esp)
# pcb is the first arg 4(%esp)
# return point is at 0(%esp)
movl 12(%esp), %eax
movl 8(%esp), %ebx
movl %esi, -4(%esp)
movl %ebp, -8(%esp)
movl 4(%esp), %esi
movl 0(%esi), %ebp # allocation pointer is at 0(pcb)
2006-11-23 19:53:15 -05:00
subl $16, %esp # 24 for alignment
2006-11-23 19:38:26 -05:00
movl %esp, 24(%esi) # save esp in pcb->system_stack
movl %ebx, %esp # load scheme stack from second arg
cmpl $-4, %eax
jne L_multi_reentry
movl -4(%esp), %eax
ret
L_multi_reentry:
movl 0(%esp), %ebx
jmp *-9(%ebx)
2006-11-23 19:42:39 -05:00
.align 8
2006-11-30 18:12:01 -05:00
ik_foreign_call:
2006-11-23 19:53:15 -05:00
_ik_foreign_call:
2006-11-23 19:42:39 -05:00
movl %esp, 8(%esi) # (movl fpr (pcb-ref 'frame-pointer))
movl %ebp, 0(%esi) # (movl apr (pcb-ref 'allocation-pointer))
movl %esp, %ebx # (movl fpr ebx)
movl 24(%esi), %esp # (movl (pcb-ref 'system-stack) esp)
2006-11-23 19:53:15 -05:00
# %esp is the system stack, %eax is the index to the last arg,
# %esi is the pcb.
# Now, the value of %esp is 16-byte aligned
# we always push %esi (4 bytes) and do a call (4 bytes),
# 0 args require 6 (2) pushes => argc= 0 (0000): %esp += -8
# 1 args require 5 (1) pushes => argc= -4 (1100): %esp += -4
# 2 args require 4 (0) pushes => argc= -8 (1000): %esp += 0
# 3 args require 3 (3) pushes => argc= -12 (0100): %esp += -12
movl %eax, %ecx
andl $15, %ecx
check_ecx:
cmpl $8, %ecx
je L_zero
cmpl $12, %ecx
je L_one
cmpl $0, %ecx
je L_two
pushl $0
L_two:
pushl $0
L_one:
pushl $0
L_zero:
2006-11-23 19:42:39 -05:00
pushl %esi # (pushl pcr)
cmpl $0, %eax # (cmpl (int 0) eax)
je L_set # (je (label Lset))
L_loop: # (label Lloop)
movl (%ebx,%eax), %ecx # (movl (mem ebx eax) ecx)
pushl %ecx # (pushl ecx)
addl $4, %eax # (addl (int 4) eax)
cmpl $0, %eax # (cmpl (int 0) eax)
jne L_loop # (jne (label Lloop))
L_set: # (label Lset)
call *%edi # (call cpr)
movl 8(%esi), %esp # (movl (pcb-ref 'frame-pointer) fpr)
movl 0(%esi), %ebp # (movl (pcb-ref 'allocation-pointer) apr)
ret # (ret)))