| ! Copyright (C) 2001, 2002, 2003, 2004, 2005, 2006, 2007 |
| ! Imagination Technologies Ltd |
| ! |
| ! Integer divide routines. |
| ! |
| |
| .text |
| .global ___udivsi3 |
| .type ___udivsi3,function |
| .align 2 |
| ___udivsi3: |
| !! |
| !! Since core is signed divide case, just set control variable |
| !! |
| MOV D1Re0,D0Ar2 ! Au already in A1Ar1, Bu -> D1Re0 |
| MOV D0Re0,#0 ! Result is 0 |
| MOV D0Ar4,#0 ! Return positive result |
| B $LIDMCUStart |
| .size ___udivsi3,.-___udivsi3 |
| |
| !! |
| !! 32-bit division signed i/p - passed signed 32-bit numbers |
| !! |
| .global ___divsi3 |
| .type ___divsi3,function |
| .align 2 |
| ___divsi3: |
| !! |
| !! A already in D1Ar1, B already in D0Ar2 -> make B abs(B) |
| !! |
| MOV D1Re0,D0Ar2 ! A already in A1Ar1, B -> D1Re0 |
| MOV D0Re0,#0 ! Result is 0 |
| XOR D0Ar4,D1Ar1,D1Re0 ! D0Ar4 -ive if result is -ive |
| ABS D1Ar1,D1Ar1 ! abs(A) -> Au |
| ABS D1Re0,D1Re0 ! abs(B) -> Bu |
| $LIDMCUStart: |
| CMP D1Ar1,D1Re0 ! Is ( Au > Bu )? |
| LSR D1Ar3,D1Ar1,#2 ! Calculate (Au & (~3)) >> 2 |
| CMPHI D1Re0,D1Ar3 ! OR ( (Au & (~3)) <= (Bu << 2) )? |
| LSLSHI D1Ar3,D1Re0,#1 ! Buq = Bu << 1 |
| BLS $LIDMCUSetup ! Yes: Do normal divide |
| !! |
| !! Quick divide setup can assume that CurBit only needs to start at 2 |
| !! |
| $LIDMCQuick: |
| CMP D1Ar1,D1Ar3 ! ( A >= Buq )? |
| ADDCC D0Re0,D0Re0,#2 ! If yes result += 2 |
| SUBCC D1Ar1,D1Ar1,D1Ar3 ! and A -= Buq |
| CMP D1Ar1,D1Re0 ! ( A >= Bu )? |
| ADDCC D0Re0,D0Re0,#1 ! If yes result += 1 |
| SUBCC D1Ar1,D1Ar1,D1Re0 ! and A -= Bu |
| ORS D0Ar4,D0Ar4,D0Ar4 ! Return neg result? |
| NEG D0Ar2,D0Re0 ! Calulate neg result |
| MOVMI D0Re0,D0Ar2 ! Yes: Take neg result |
| $LIDMCRet: |
| MOV PC,D1RtP |
| !! |
| !! Setup for general unsigned divide code |
| !! |
| !! D0Re0 is used to form the result, already set to Zero |
| !! D1Re0 is the input Bu value, this gets trashed |
| !! D0Ar6 is curbit which is set to 1 at the start and shifted up |
| !! D0Ar4 is negative if we should return a negative result |
| !! D1Ar1 is the input Au value, eventually this holds the remainder |
| !! |
| $LIDMCUSetup: |
| CMP D1Ar1,D1Re0 ! Is ( Au < Bu )? |
| MOV D0Ar6,#1 ! Set curbit to 1 |
| BCS $LIDMCRet ! Yes: Return 0 remainder Au |
| !! |
| !! Calculate alignment using FFB instruction |
| !! |
| FFB D1Ar5,D1Ar1 ! Find first bit of Au |
| ANDN D1Ar5,D1Ar5,#31 ! Handle exceptional case. |
| ORN D1Ar5,D1Ar5,#31 ! if N bit set, set to 31 |
| FFB D1Ar3,D1Re0 ! Find first bit of Bu |
| ANDN D1Ar3,D1Ar3,#31 ! Handle exceptional case. |
| ORN D1Ar3,D1Ar3,#31 ! if N bit set, set to 31 |
| SUBS D1Ar3,D1Ar5,D1Ar3 ! calculate diff, ffbA - ffbB |
| MOV D0Ar2,D1Ar3 ! copy into bank 0 |
| LSLGT D1Re0,D1Re0,D1Ar3 ! ( > 0) ? left shift B |
| LSLGT D0Ar6,D0Ar6,D0Ar2 ! ( > 0) ? left shift curbit |
| !! |
| !! Now we start the divide proper, logic is |
| !! |
| !! if ( A >= B ) add curbit to result and subtract B from A |
| !! shift curbit and B down by 1 in either case |
| !! |
| $LIDMCLoop: |
| CMP D1Ar1, D1Re0 ! ( A >= B )? |
| ADDCC D0Re0, D0Re0, D0Ar6 ! If yes result += curbit |
| SUBCC D1Ar1, D1Ar1, D1Re0 ! and A -= B |
| LSRS D0Ar6, D0Ar6, #1 ! Shift down curbit, is it zero? |
| LSR D1Re0, D1Re0, #1 ! Shift down B |
| BNZ $LIDMCLoop ! Was single bit in curbit lost? |
| ORS D0Ar4,D0Ar4,D0Ar4 ! Return neg result? |
| NEG D0Ar2,D0Re0 ! Calulate neg result |
| MOVMI D0Re0,D0Ar2 ! Yes: Take neg result |
| MOV PC,D1RtP |
| .size ___divsi3,.-___divsi3 |