252 lines
8.6 KiB
OCaml
252 lines
8.6 KiB
OCaml
(***********************************************************************)
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(* *)
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(* Objective Caml *)
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(* *)
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(* Xavier Leroy, projet Cristal, INRIA Rocquencourt *)
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(* *)
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(* Copyright 1996 Institut National de Recherche en Informatique et *)
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(* Automatique. Distributed only by permission. *)
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(* *)
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(***********************************************************************)
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(* $Id$ *)
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(* Instruction scheduling *)
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open Misc
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open Reg
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open Mach
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open Linearize
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(* Determine whether an instruction ends a basic block or not *)
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let in_basic_block instr =
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match instr.desc with
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Lop op ->
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begin match op with
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Icall_ind | Icall_imm _ | Itailcall_ind | Itailcall_imm _ -> false
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| Iextcall(_, _) -> false
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| Istackoffset _ -> false
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| Istore(_, _) -> false
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| Ialloc _ -> false
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| op -> Proc.oper_latency op >= 0
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(* The processor description can return a latency of -1 to signal
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a specific instruction that terminates a basic block, e.g.
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Istore_symbol for the I386. *)
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end
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| Lreloadretaddr -> true
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| _ -> false
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(* Estimate the delay needed to evaluate an instruction. *)
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let reload_retaddr_latency =
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Proc.oper_latency (Iload(Cmm.Word, Arch.identity_addressing))
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let instr_latency instr =
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match instr.desc with
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Lop op -> Proc.oper_latency op
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| Lreloadretaddr -> reload_retaddr_latency
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| _ -> fatal_error "Scheduling.instr_latency"
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(* Representation of the code DAG. *)
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type code_dag_node =
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{ instr: instruction; (* The instruction *)
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delay: int; (* How many cycles it needs *)
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mutable sons: (code_dag_node * int) list;
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(* Instructions that depend on it *)
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mutable date: int; (* Start date *)
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mutable length: int; (* Length of longest path to result *)
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mutable ancestors: int; (* Number of ancestors *)
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mutable emitted_ancestors: int } (* Number of emitted ancestors *)
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let dummy_node =
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{ instr = end_instr; delay = 0; sons = []; date = 0;
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length = -1; ancestors = 0; emitted_ancestors = 0 }
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(* The code dag itself is represented by two tables from registers to nodes:
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- "results" maps registers to the instructions that produced them;
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- "uses" maps registers to the instructions that use them. *)
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let code_results = (Hashtbl.create 31 : (location, code_dag_node) Hashtbl.t)
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let code_uses = (Hashtbl.create 31 : (location, code_dag_node) Hashtbl.t)
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let clear_code_dag () =
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Hashtbl.clear code_results;
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Hashtbl.clear code_uses
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(* Add an instruction to the code DAG *)
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let add_edge ancestor son delay =
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ancestor.sons <- (son, delay) :: ancestor.sons;
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son.ancestors <- son.ancestors + 1
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let add_instruction ready_queue instr =
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let delay = instr_latency instr in
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let node =
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{ instr = instr;
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delay = delay;
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sons = [];
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date = 0;
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length = -1;
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ancestors = 0;
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emitted_ancestors = 0 } in
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(* Add edges from all instructions that define one of the registers used *)
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for i = 0 to Array.length instr.arg - 1 do
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try
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let ancestor = Hashtbl.find code_results instr.arg.(i).loc in
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add_edge ancestor node ancestor.delay
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with Not_found ->
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()
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done;
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(* Also add edges from all instructions that use one of the results
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of this instruction, so that evaluation order is preserved. *)
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for i = 0 to Array.length instr.res - 1 do
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let ancestors = Hashtbl.find_all code_uses instr.res.(i).loc in
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List.iter (fun ancestor -> add_edge ancestor node 0) ancestors
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done;
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(* Also add edges from all instructions that have already defined one
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of the results of this instruction, so that evaluation order
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is preserved. *)
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for i = 0 to Array.length instr.res - 1 do
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try
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let ancestor = Hashtbl.find code_results instr.res.(i).loc in
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add_edge ancestor node 0
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with Not_found ->
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()
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done;
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(* Remember the registers used and produced by this instruction *)
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for i = 0 to Array.length instr.res - 1 do
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Hashtbl.add code_results instr.res.(i).loc node
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done;
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for i = 0 to Array.length instr.arg - 1 do
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Hashtbl.add code_uses instr.arg.(i).loc node
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done;
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(* If this is a root instruction (all arguments already computed),
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add it to the ready queue *)
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if node.ancestors = 0 then node :: ready_queue else ready_queue
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(* Compute length of longest path to a result.
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For leafs of the DAG, see whether their result is used in the instruction
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immediately following the basic block (a "critical" output). *)
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let is_critical critical_outputs results =
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try
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for i = 0 to Array.length results - 1 do
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let r = results.(i).loc in
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for j = 0 to Array.length critical_outputs - 1 do
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if critical_outputs.(j).loc = r then raise Exit
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done
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done;
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false
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with Exit ->
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true
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let rec longest_path critical_outputs node =
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if node.length < 0 then begin
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match node.sons with
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[] ->
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node.length <-
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if is_critical critical_outputs node.instr.res
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or node.instr.desc = Lreloadretaddr (* alway critical *)
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then node.delay
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else 0
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| sons ->
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node.length <-
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List.fold_left
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(fun len (son, delay) ->
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max len (longest_path critical_outputs son + delay))
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0 sons
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end;
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node.length
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(* Given a list of instructions with estimated start date, choose one
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that we can start (start date <= current date) and that has
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maximal distance to result. If we can't find any, return None. *)
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let extract_ready_instr date queue =
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let rec extract best = function
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[] ->
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if best == dummy_node then None else Some best
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| instr :: rem ->
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let new_best =
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if instr.date <= date & instr.length > best.length
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then instr else best in
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extract new_best rem in
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extract dummy_node queue
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(* Remove an instruction from the ready queue *)
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let rec remove_instr node = function
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[] -> []
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| instr :: rem ->
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if instr == node then rem else instr :: remove_instr node rem
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(* Schedule a basic block, adding its instructions in front of the given
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instruction sequence *)
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let rec reschedule ready_queue date cont =
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match ready_queue with
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[] -> cont
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| _ ->
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(* Find "most ready" instruction in queue *)
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match extract_ready_instr date ready_queue with
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None ->
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(* Try again, one cycle later *)
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reschedule ready_queue (date + 1) cont
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| Some node ->
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(* Update the start date and number of ancestors emitted of
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all descendents of this node. Enter those that become ready
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in the queue. *)
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let new_queue = ref (remove_instr node ready_queue) in
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List.iter
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(fun (son, delay) ->
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let completion_date = date + delay in
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if son.date < completion_date then son.date <- completion_date;
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son.emitted_ancestors <- son.emitted_ancestors + 1;
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if son.emitted_ancestors = son.ancestors then
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new_queue := son :: !new_queue)
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node.sons;
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instr_cons node.instr.desc node.instr.arg node.instr.res
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(reschedule !new_queue (date + 1) cont)
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(* Schedule basic blocks in an instruction sequence *)
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let rec schedule i =
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match i.desc with
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Lend -> i
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| _ ->
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if in_basic_block i then begin
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clear_code_dag();
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schedule_block [] i
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end else
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{ desc = i.desc; arg = i.arg; res = i.res; live = i.live;
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next = schedule i.next }
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and schedule_block ready_queue i =
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if in_basic_block i then
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schedule_block (add_instruction ready_queue i) i.next
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else begin
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let critical_outputs =
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match i.desc with
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Lop(Icall_ind | Itailcall_ind) -> [| i.arg.(0) |]
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| Lop(Icall_imm _ | Itailcall_imm _ | Iextcall(_, _)) -> [||]
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| Lreturn -> [||]
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| _ -> i.arg in
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List.iter (longest_path critical_outputs) ready_queue;
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reschedule ready_queue 0 (schedule i)
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end
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(* Entry point *)
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(* Don't bother to schedule for initialization code and the like. *)
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let fundecl f =
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if Proc.need_scheduling & f.fun_fast then begin
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let new_body = schedule f.fun_body in
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clear_code_dag();
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{ fun_name = f.fun_name;
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fun_body = new_body;
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fun_fast = f.fun_fast }
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end else
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f
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