Filed PR jit/66812 for the code generation issue
Dibyendu Majumdar
mobile@majumdar.org.uk
Thu Jan 1 00:00:00 GMT 2015
Apologies I think the previous description of the flow was incorrect.
Here is my second attempt (this is doing my head in so I will stop
now):
1 [2] LOADNIL 0 0
entry:
cl = (struct ravi_LClosure *)L->ci->func->value_.gc;
(void)raviV_op_loadnil (L->ci, (int)0, (int)0);
Above sets register 0 (local variable IX) to NIL
2 [3] LOADK 1 -1 ; 10
(&L->ci->u.l.base[(int)1])->value_.i = (&cl->p->k[(int)0])->value_.i;
(&L->ci->u.l.base[(int)1])->tt_ = (&cl->p->k[(int)0])->tt_;
This sets the register at 1 (temporary) to 10.
3 [3] TEST 1 1
4 [3] JMP 0 0 ; to 5
The above two bytecodes go together. This says that if register 1 is
true then do JMP else skip JMP.
So we get:
comparison_0_5 = (&L->ci->u.l.base[(int)1])->tt_ == (int)0;
comparison_0_6 = (&L->ci->u.l.base[(int)1])->tt_ == (int)1;
comparison_0_7 = (&L->ci->u.l.base[(int)1])->value_.b == (int)0;
isfalse_0_4 = comparison_0_5 || comparison_0_6 && comparison_0_7;
if (!(isfalse_0_4)) goto OP_TEST_do_jmp_2_8; else goto OP_TEST_do_skip_2_9;
Since register 1 is true (ie. not false) we therefore jump to jmp_2_8.
OP_TEST_do_jmp_2_8:
goto jmp_5_1;
jmp_5_1:
5 [3] LOADBOOL 1 0 0
This is setting register 1 to false so we get:
(&L->ci->u.l.base[(int)1])->value_.b = (int)0;
(&L->ci->u.l.base[(int)1])->tt_ = (int)1;
Next instruction is:
6 [3] TEST 1 0
7 [3] JMP 0 1 ; to 9
Now this is saying that if register 1 is false then do JMP else skip JMP.
comparison_0_11 = (&L->ci->u.l.base[(int)1])->tt_ == (int)0;
comparison_0_12 = (&L->ci->u.l.base[(int)1])->tt_ == (int)1;
comparison_0_13 = (&L->ci->u.l.base[(int)1])->value_.b == (int)0;
isfalse_0_10 = comparison_0_11 || comparison_0_12 && comparison_0_13;
if (!(!(isfalse_0_10))) goto OP_TEST_do_jmp_5_14; else goto
OP_TEST_do_skip_5_15;
Since we know that register 1 is false as above then the code should
do the jump to jmp_5_14:
OP_TEST_do_jmp_5_14:
goto jmp_9_2;
Next we jump to 9_2 which is the start of the return statement.
Here we essentially have a repeat of what was in entry block:
9 [3] LOADK 1 -1 ; 10
10 [3] TEST 1 1
11 [3] JMP 0 0 ; to 12
jmp_9_2:
(&L->ci->u.l.base[(int)1])->value_.i = (&cl->p->k[(int)0])->value_.i;
(&L->ci->u.l.base[(int)1])->tt_ = (&cl->p->k[(int)0])->tt_;
comparison_0_17 = (&L->ci->u.l.base[(int)1])->tt_ == (int)0;
comparison_0_18 = (&L->ci->u.l.base[(int)1])->tt_ == (int)1;
comparison_0_19 = (&L->ci->u.l.base[(int)1])->value_.b == (int)0;
isfalse_0_16 = comparison_0_17 || comparison_0_18 && comparison_0_19;
if (!(isfalse_0_16)) goto OP_TEST_do_jmp_9_20; else goto OP_TEST_do_skip_9_21;
So the JMP is taken and we go to 9_20:
OP_TEST_do_jmp_9_20:
goto jmp_12_3;
jmp_12_3:
12 [3] LOADBOOL 1 0 0
(&L->ci->u.l.base[(int)1])->value_.b = (int)0;
(&L->ci->u.l.base[(int)1])->tt_ = (int)1;
Above sets the register 1 to false and this is the return value.
Next the return instruction begins:
L->top = &L->ci->u.l.base[(int)2];
comparison_0_22 = cl->p->sizep > (int)0;
if (comparison_0_22) goto OP_RETURN_if_sizep_gt_0_12_23; else goto
OP_RETURN_else_sizep_gt_0_12_24;
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