Lines Matching defs:shift
701 Int shift = 0;
711 case 1: shift = 0; break;
712 case 4: shift = 2; break;
713 case 8: shift = 3; break;
722 ... base(%ebp, %tmp, shift) ...
734 X86AMode_IRRS( descr->base, hregX86_EBP(), tmp, shift );
978 /* Perhaps a shift op? */
1017 /* Now consider the shift amount. If it's a literal, we
1081 Int shift = (e->Iex.Binop.op == Iop_MullS8
1090 addInstr(env, X86Instr_Sh32(Xsh_SHL, shift, a16));
1091 addInstr(env, X86Instr_Sh32(Xsh_SHL, shift, b16));
1092 addInstr(env, X86Instr_Sh32(shr_op, shift, a16));
1093 addInstr(env, X86Instr_Sh32(shr_op, shift, b16));
1103 /* shift this right 8 bits so as to conform to CmpF64
1290 Int shift = e->Iex.Unop.op == Iop_16HIto8 ? 8 : 16;
1292 addInstr(env, X86Instr_Sh32(Xsh_SHR, shift, dst));
1575 UInt shift = e->Iex.Binop.arg1
1578 if (shift == 1 || shift == 2 || shift == 3) {
1582 return X86AMode_IRRS(imm32, r1, r2, shift);
1593 UInt shift = e->Iex.Binop.arg2->Iex.Binop.arg2->Iex.Const.con->Ico.U8;
1594 if (shift == 1 || shift == 2 || shift == 3) {
1597 return X86AMode_IRRS(0, r1, r2, shift);
2299 to be shifted into %hi:%lo, and the shift amount into
2303 -- shift amt %cl % 32
2305 -- shift amt %cl % 32
2307 Now, if (shift amount % 64) is in the range 32 .. 63,
2326 /* Ok. Now shift amt is in %ecx, and value is in tHi/tLo
2341 to be shifted into %hi:%lo, and the shift amount into
2345 -- shift amt %cl % 32
2347 -- shift amt %cl % 32
2349 Now, if (shift amount % 64) is in the range 32 .. 63,
2368 /* Ok. Now shift amt is in %ecx, and value is in tHi/tLo
