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