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/* SPDX-License-Identifier: GPL-2.0-or-later */
/*
* Copyright (C) 2003-2013 Altera Corporation
* All rights reserved.
*/
#include <linux/linkage.h>
#include <asm/entry.h>
.set noat
.set nobreak
/*
* Explicitly allow the use of r1 (the assembler temporary register)
* within this code. This register is normally reserved for the use of
* the compiler.
*/
ENTRY(instruction_trap)
ldw r1, PT_R1(sp) // Restore registers
ldw r2, PT_R2(sp)
ldw r3, PT_R3(sp)
ldw r4, PT_R4(sp)
ldw r5, PT_R5(sp)
ldw r6, PT_R6(sp)
ldw r7, PT_R7(sp)
ldw r8, PT_R8(sp)
ldw r9, PT_R9(sp)
ldw r10, PT_R10(sp)
ldw r11, PT_R11(sp)
ldw r12, PT_R12(sp)
ldw r13, PT_R13(sp)
ldw r14, PT_R14(sp)
ldw r15, PT_R15(sp)
ldw ra, PT_RA(sp)
ldw fp, PT_FP(sp)
ldw gp, PT_GP(sp)
ldw et, PT_ESTATUS(sp)
wrctl estatus, et
ldw ea, PT_EA(sp)
ldw et, PT_SP(sp) /* backup sp in et */
addi sp, sp, PT_REGS_SIZE
/* INSTRUCTION EMULATION
* ---------------------
*
* Nios II processors generate exceptions for unimplemented instructions.
* The routines below emulate these instructions. Depending on the
* processor core, the only instructions that might need to be emulated
* are div, divu, mul, muli, mulxss, mulxsu, and mulxuu.
*
* The emulations match the instructions, except for the following
* limitations:
*
* 1) The emulation routines do not emulate the use of the exception
* temporary register (et) as a source operand because the exception
* handler already has modified it.
*
* 2) The routines do not emulate the use of the stack pointer (sp) or
* the exception return address register (ea) as a destination because
* modifying these registers crashes the exception handler or the
* interrupted routine.
*
* Detailed Design
* ---------------
*
* The emulation routines expect the contents of integer registers r0-r31
* to be on the stack at addresses sp, 4(sp), 8(sp), ... 124(sp). The
* routines retrieve source operands from the stack and modify the
* destination register's value on the stack prior to the end of the
* exception handler. Then all registers except the destination register
* are restored to their previous values.
*
* The instruction that causes the exception is found at address -4(ea).
* The instruction's OP and OPX fields identify the operation to be
* performed.
*
* One instruction, muli, is an I-type instruction that is identified by
* an OP field of 0x24.
*
* muli AAAAA,BBBBB,IIIIIIIIIIIIIIII,-0x24-
* 27 22 6 0 <-- LSB of field
*
* The remaining emulated instructions are R-type and have an OP field
* of 0x3a. Their OPX fields identify them.
*
* R-type AAAAA,BBBBB,CCCCC,XXXXXX,NNNNN,-0x3a-
* 27 22 17 11 6 0 <-- LSB of field
*
*
* Opcode Encoding. muli is identified by its OP value. Then OPX & 0x02
* is used to differentiate between the division opcodes and the
* remaining multiplication opcodes.
*
* Instruction OP OPX OPX & 0x02
* ----------- ---- ---- ----------
* muli 0x24
* divu 0x3a 0x24 0
* div 0x3a 0x25 0
* mul 0x3a 0x27 != 0
* mulxuu 0x3a 0x07 != 0
* mulxsu 0x3a 0x17 != 0
* mulxss 0x3a 0x1f != 0
*/
/*
* Save everything on the stack to make it easy for the emulation
* routines to retrieve the source register operands.
*/
addi sp, sp, -128
stw zero, 0(sp) /* Save zero on stack to avoid special case for r0. */
stw r1, 4(sp)
stw r2, 8(sp)
stw r3, 12(sp)
stw r4, 16(sp)
stw r5, 20(sp)
stw r6, 24(sp)
stw r7, 28(sp)
stw r8, 32(sp)
stw r9, 36(sp)
stw r10, 40(sp)
stw r11, 44(sp)
stw r12, 48(sp)
stw r13, 52(sp)
stw r14, 56(sp)
stw r15, 60(sp)
stw r16, 64(sp)
stw r17, 68(sp)
stw r18, 72(sp)
stw r19, 76(sp)
stw r20, 80(sp)
stw r21, 84(sp)
stw r22, 88(sp)
stw r23, 92(sp)
/* Don't bother to save et. It's already been changed. */
rdctl r5, estatus
stw r5, 100(sp)
stw gp, 104(sp)
stw et, 108(sp) /* et contains previous sp value. */
stw fp, 112(sp)
stw ea, 116(sp)
stw ra, 120(sp)
/*
* Split the instruction into its fields. We need 4*A, 4*B, and 4*C as
* offsets to the stack pointer for access to the stored register values.
*/
ldw r2,-4(ea) /* r2 = AAAAA,BBBBB,IIIIIIIIIIIIIIII,PPPPPP */
roli r3, r2, 7 /* r3 = BBB,IIIIIIIIIIIIIIII,PPPPPP,AAAAA,BB */
roli r4, r3, 3 /* r4 = IIIIIIIIIIIIIIII,PPPPPP,AAAAA,BBBBB */
roli r5, r4, 2 /* r5 = IIIIIIIIIIIIII,PPPPPP,AAAAA,BBBBB,II */
srai r4, r4, 16 /* r4 = (sign-extended) IMM16 */
roli r6, r5, 5 /* r6 = XXXX,NNNNN,PPPPPP,AAAAA,BBBBB,CCCCC,XX */
andi r2, r2, 0x3f /* r2 = 00000000000000000000000000,PPPPPP */
andi r3, r3, 0x7c /* r3 = 0000000000000000000000000,AAAAA,00 */
andi r5, r5, 0x7c /* r5 = 0000000000000000000000000,BBBBB,00 */
andi r6, r6, 0x7c /* r6 = 0000000000000000000000000,CCCCC,00 */
/* Now
* r2 = OP
* r3 = 4*A
* r4 = IMM16 (sign extended)
* r5 = 4*B
* r6 = 4*C
*/
/*
* Get the operands.
*
* It is necessary to check for muli because it uses an I-type
* instruction format, while the other instructions are have an R-type
* format.
*
* Prepare for either multiplication or division loop.
* They both loop 32 times.
*/
movi r14, 32
add r3, r3, sp /* r3 = address of A-operand. */
ldw r3, 0(r3) /* r3 = A-operand. */
movi r7, 0x24 /* muli opcode (I-type instruction format) */
beq r2, r7, mul_immed /* muli doesn't use the B register as a source */
add r5, r5, sp /* r5 = address of B-operand. */
ldw r5, 0(r5) /* r5 = B-operand. */
/* r4 = SSSSSSSSSSSSSSSS,-----IMM16------ */
/* IMM16 not needed, align OPX portion */
/* r4 = SSSSSSSSSSSSSSSS,CCCCC,-OPX--,00000 */
srli r4, r4, 5 /* r4 = 00000,SSSSSSSSSSSSSSSS,CCCCC,-OPX-- */
andi r4, r4, 0x3f /* r4 = 00000000000000000000000000,-OPX-- */
/* Now
* r2 = OP
* r3 = src1
* r5 = src2
* r4 = OPX (no longer can be muli)
* r6 = 4*C
*/
/*
* Multiply or Divide?
*/
andi r7, r4, 0x02 /* For R-type multiply instructions,
OPX & 0x02 != 0 */
bne r7, zero, multiply
/* DIVISION
*
* Divide an unsigned dividend by an unsigned divisor using
* a shift-and-subtract algorithm. The example below shows
* 43 div 7 = 6 for 8-bit integers. This classic algorithm uses a
* single register to store both the dividend and the quotient,
* allowing both values to be shifted with a single instruction.
*
* remainder dividend:quotient
* --------- -----------------
* initialize 00000000 00101011:
* shift 00000000 0101011:_
* remainder >= divisor? no 00000000 0101011:0
* shift 00000000 101011:0_
* remainder >= divisor? no 00000000 101011:00
* shift 00000001 01011:00_
* remainder >= divisor? no 00000001 01011:000
* shift 00000010 1011:000_
* remainder >= divisor? no 00000010 1011:0000
* shift 00000101 011:0000_
* remainder >= divisor? no 00000101 011:00000
* shift 00001010 11:00000_
* remainder >= divisor? yes 00001010 11:000001
* remainder -= divisor - 00000111
* ----------
* 00000011 11:000001
* shift 00000111 1:000001_
* remainder >= divisor? yes 00000111 1:0000011
* remainder -= divisor - 00000111
* ----------
* 00000000 1:0000011
* shift 00000001 :0000011_
* remainder >= divisor? no 00000001 :00000110
*
* The quotient is 00000110.
*/
divide:
/*
* Prepare for division by assuming the result
* is unsigned, and storing its "sign" as 0.
*/
movi r17, 0
/* Which division opcode? */
xori r7, r4, 0x25 /* OPX of div */
bne r7, zero, unsigned_division
/*
* OPX is div. Determine and store the sign of the quotient.
* Then take the absolute value of both operands.
*/
xor r17, r3, r5 /* MSB contains sign of quotient */
bge r3,zero,dividend_is_nonnegative
sub r3, zero, r3 /* -r3 */
dividend_is_nonnegative:
bge r5, zero, divisor_is_nonnegative
sub r5, zero, r5 /* -r5 */
divisor_is_nonnegative:
unsigned_division:
/* Initialize the unsigned-division loop. */
movi r13, 0 /* remainder = 0 */
/* Now
* r3 = dividend : quotient
* r4 = 0x25 for div, 0x24 for divu
* r5 = divisor
* r13 = remainder
* r14 = loop counter (already initialized to 32)
* r17 = MSB contains sign of quotient
*/
/*
* for (count = 32; count > 0; --count)
* {
*/
divide_loop:
/*
* Division:
*
* (remainder:dividend:quotient) <<= 1;
*/
slli r13, r13, 1
cmplt r7, r3, zero /* r7 = MSB of r3 */
or r13, r13, r7
slli r3, r3, 1
/*
* if (remainder >= divisor)
* {
* set LSB of quotient
* remainder -= divisor;
* }
*/
bltu r13, r5, div_skip
ori r3, r3, 1
sub r13, r13, r5
div_skip:
/*
* }
*/
subi r14, r
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