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;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;
; Filename: test.S
;
; Project: Zip CPU -- a small, lightweight, RISC CPU soft core
;
; Purpose: A disorganized test, just showing some initial operation of
; the CPU. As a disorganized test, it doesn't prove anything
; beyond the generic operation of the CPU.
;
; Status: As of August, 2015, this file assembles, builds, and passes
; all of its tests in the Verilator simulator.
;
; Okay, as of 15 August, there are now some tests that don't pass.
; In particular, the #include test used to pass but didn't pass today.
; Likewise the PUSH() macro test hasn't passed yet. Finally, be aware
; that this implementation is specific to where it loads on a board.
; I tried loading it on my Basys development board, where I had placed
; RAM in a different location and ... things didn't work out so well.
; So grep the __here__ line and adjust it for where you intend to load
; this file.
;
; In general, as I'm building the CPU, I'm modifying this file to place
; more and more capability tests within the file. If the Lord is
; willing, this will become the proof that the CPU completely works.
;
;
; Creator: Dan Gisselquist, Ph.D.
; Gisselquist Technology, LLC
;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;
; Copyright (C) 2015, Gisselquist Technology, LLC
;
; This program is free software (firmware): you can redistribute it and/or
; modify it under the terms of the GNU General Public License as published
; by the Free Software Foundation, either version 3 of the License, or (at
; your option) any later version.
;
; This program is distributed in the hope that it will be useful, but WITHOUT
; ANY WARRANTY; without even the implied warranty of MERCHANTIBILITY or
; FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
; for more details.
;
; License: GPL, v3, as defined and found on www.gnu.org,
; http://www.gnu.org/licenses/gpl.html
;
;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;
#include "sys.i"
sys.bus equ 0xc0000000
sys.breaken equ 0x080
sys.step equ 0x040
sys.gie equ 0x020
sys.sleep equ 0x010
sys.ccv equ 0x008
sys.ccn equ 0x004
sys.ccc equ 0x002
sys.ccz equ 0x001
sys.cctrap equ 0x200
sys.bu.pic equ 0x000
sys.bus.wdt equ 0x001
sys.bus.cache equ 0x002
sys.bus.ctrpic equ 0x003
sys.bus.tma equ 0x004
sys.bus.tmb equ 0x005
sys.bus.tmc equ 0x006
sys.bus.jiffies equ 0x007
sys.bus.mtask equ 0x008
sys.bus.mpstl equ 0x009
sys.bus.mastl equ 0x00a
sys.bus.mstl equ 0x00b
sys.bus.utask equ 0x00c
sys.bus.upstl equ 0x00d
sys.bus.uastl equ 0x00e
sys.bus.ustl equ 0x00f
#define DO_TEST_ASSEMBLER
#define LJMP_TEST
#define EARLY_BRANCH_TEST
#define BREAK_TEST
#define OVERFLOW_TEST
#define CARRY_TEST
#define LOOP_TEST
#define SHIFT_TEST
#define TRAP_TEST
#define MPY_TEST
#define PUSH_TEST
#define PIPELINE_STACK_TEST
#define MEM_PIPELINE_TEST
#define CONDITIONAL_EXECUTION_TEST
#define NOWAIT_PIPELINE_TEST // Were wait states btwn regs removed properly?
#define BCMEM_TEST // Do memory and conditions work well together?
test:
#ifdef DO_TEST_ASSEMBLER
; We start out by testing our assembler. We give it some instructions, which
; are then manually checked by disassembling/dumping the result and making
; certain they match. This is not an automated test, but it is an important
; one.
noop
bra continue_test_with_testable_instructions
break
wait
break
busy
rtu
continue_test_with_testable_instructions:
; Now, let's place the assembler into a known state
clr r0
clr r1
clr r2
clr r3
clr r4
clr r5
clr r6
clr r7
clr r8
clr r9
clr r10
clr r11
clr r12
clr r13
; Don't clear the CC register
; Don't clear the SP register
; And repeat for the user registers
mov R0,uR0
mov R0,uR1
mov R0,uR2
mov R0,uR3
mov R0,uR4
mov R0,uR5
mov R0,uR6
mov R0,uR7
mov R0,uR8
mov R0,uR9
mov R0,uR10
mov R0,uR11
mov R0,uR12
mov R0,uR13
mov R0,uCC
; Don't clear the user PC register
; Now, let's try loading some constants into registers
; Specifically, we're testing the LDI, LDIHI, and LDILO instructions
dead_beef equ 0xdeadbeef
ldi 0x0dead,r5
ldi 0x0beef,r6
ldi 0xdeadbeef,r7
ldihi 0xdead, r8
ldilo 0xbeef, r8
ldi dead_beef,r9
cmp r5,r6
bz test_failure
cmp r7,r8
bnz test_failure
ldi $deadbeefh,r7 ; Try loading with the $[HEX]h mneumonic
cmp r7,r8
bnz test_failure
cmp r7,r9
bnz test_failure
bra skip_dead_beef
dead_beef.base:
word 0
fill 5,dead_beef
word 0
dead_beef.zero equ 0
dead_beef.values equ 1
skip_dead_beef:
lod dead_beef.base(pc),r10 ; Should load a zero here
cmp r10,r11 ; r11 should still be zero from init abv
bnz test_failure
mov dead_beef.base(pc),r10 ; Now, let's get the address
lod dead_beef.values(r10),r10 ; r10 now equals 0xdeadbeef
cmp r10,r9
bnz test_failure
; Test whether or not our operator precedence rules work
ldi 5+3*8,r0
ldi 3*8+5,r1
cmp r0,r1
bnz test_failure
ldi (5+3)*8,r0
ldi 8*(3+5),r1
cmp r0,r1
bnz test_failure
; Test whether or not we can properly decode OCTAL values
clr r0 ; Re-clear our register set first
clr r1
clr r2
clr r3
clr r4
clr r5
clr r6
clr r7
clr r8
clr r9
clr r10
clr r11
clr r12
clr r13
;
ldi $024o,r0
ldi $20,r1
cmp r0,r1
bnz test_failure
ldi $024,r0
cmp r0,r1
bnz test_failure
clr r0
clr r1
mov $1+r0,r2
mov $2+r0,r3
mov $22h+r0,r4
mov $377h+r0,ur5
noop
nop
add r2,r0
add $32,r0
add $-33,r0
bnz test_failure
not r0
bge test_failure
junk_address:
clrf r0
bnz test_failure
ldi $5,r1
cmp $0+r0,r1
not.lt r0
not.ge r1
mov junk_address(pc),r2 ; Test pc-relative addressing
mov junk_address(pc),r3
cmp r2,r3
bnz test_failure
lod junk_address(pc),r5 ; Test loads with pc-relative addressing
lod junk_address(pc),r6
cmp r5,r6
bnz test_failure
#endif
#ifdef NOONE // Testing comments after ifdef
#else ; After else
#endif /* and after endif */
#ifdef LJMP_TEST
// A long jump is a 32-bit instruction followed by a 32-bit address.
// The CPU is supposed to jump to this address. At issue in this test,
// which can only really be verified by watching it in person currently,
// is how fast this branch can take place. Currently, it takes four
// clocks--not that bad.
//
// Although really long jumps, we also test some of our early branching
// forms here as well:
// 1. Add to PC
// 2. LOD (PC),PC (the long jump itself)
// 3. LDI x,PC // An early branch target not tested elsewhere
//
CLR R0
CLR R1
LJMP
.dat __here__+0x0100000+4
ADD 1,R0
ADD 1,R0
ADD 1,R0
ADD 1,R0
ADD 1,R0
ADD 1,R0
CMP 3,R0
BNZ test_failure
LOD.Z __here__+2(PC),PC
BRA test_failure
.dat __here__+0x0100000+2
ADD 1,R0
ADD 1,R0
ADD 1,R0
CMP 5,R0
BNZ test_failure
// And our last early branching test
LDI 0x0100000+__here__+4,PC
ADD 1,R0
ADD 1,R0
ADD 1,R0
SUB 1,R0
CMP 4,R0
BNZ test_failure
#endif
#ifdef EARLY_BRANCH_TEST
// Unlike the previous test, this test is going to see whether or not
// early branching messes with the pipeline.
BRA eb_a
BUSY
eb_a:
BRA eb_b
NOP
BUSY
eb_b:
BRA eb_c
NOP
NOP
BUSY
eb_c:
BRA eb_d
NOP
NOP
NOP
BUSY
eb_d:
BRA eb_e
NOP
NOP
NOP
NOP
BUSY
eb_e:
NOOP
// Only problem is, I don't expect it to mess with the pipeline unless
// the pipeline is full. Therefore we are interested in something which
// is not an early branch, conflicting with early branches. So let's
// try loading our pipeline in all kinds of different configurations,
// just to see which if the conditional branch always annihilates the
// early branch as desired.
//
CLR R0
BZ ebz_a
BUSY
ebz_a:
BZ ebz_b
NOP
BUSY
ebz_b:
BZ ebz_c
NOP
NOP
BUSY
// Let's repeat that last test, just in case the cache reloaded itself
// in the middle and we didn't get our proper test.
ebz_c:
BZ ebz_d
NOP
NOP
BUSY
ebz_d:
BZ ebz_e
NOP
NOP
NOP
BUSY
ebz_e:
BZ ebz_f
NOP
NOP
NOP
NOP
BUSY
ebz_f:
NOOP
#endif
#ifdef BREAK_TEST
breaktest:
bra breaksupervisor
breakuser:
clr r0
mov 1+r0,r1
mov 1+r1,r2
mov 1+r2,r3
break ; At address 0x0100097
mov 1+r4,r5
mov 1+r5,r6
clr cc
busy
breaksupervisor:
ldi -1,r0
mov breakuser(pc),upc
rtu ; Should just keep returning immediately
mov upc,r0
rtu
rtu
mov upc,r1
cmp r0,r1
bnz test_failure
#endif
#ifdef TRAP_TEST
traptest:
bra traptest_supervisor
busy
traptest_user:
trap 0
busy
traptest_supervisor:
mov traptest_user(pc),upc
rtu
mov ucc,r0
tst sys.cctrap,r0
tst.nz sys.gie,r0
bz test_failure
#endif
testbench:
// Let's build a software test bench.
ldi $c0000000h,r12 ; Set R12 to point to our peripheral address
mov r12,ur12
mov test_start(pc),upc
mov stack(pc),usp
ldi 0x8000ffff,r0 ; Clear interrupts, turn all vectors off
sto r0,(r12)
rtu
mov ucc,r0
CMP sys.cctrap+sys.gie,r0
bnz test_failure
halt
// Go into an infinite loop if the trap fails
// Permanent loop instruction -- a busy halt if you will
test_failure:
busy
; Now for a series of tests. If the test fails, call the trap
; interrupt with the test number that failed. Upon completion,
; call the trap with #0.
; Test LDI to PC
; Some data registers
test_data:
.dat __here__+0x0100000+5
test_start:
ldi $0x01000,r11
ldi -1,r10
lod test_data+pc,pc
clr r10
noop
cmp $0,r10
trap.z r11
add $1,r0
add $1,r0
#ifdef OVERFLOW_TEST
// Let's test whether overflow works
ldi $0x02000,r11
ldi $-1,r0
lsr $1,r0
add $1,r0
bv first_overflow_passes
trap r11
first_overflow_passes:
// Overflow set from subtraction
ldi $0x03000,r11
ldi $1,r0
rol $31,r0 ; rol $31,r0
sub $1,r0
bv subtraction_overflow_passes
trap r11
subtraction_overflow_passes:
// Overflow set from LSR
ldi $0x04000,r11
ldi $1,r0
rol $31,r0 ; rol $31,r0
lsr $1,r0
bv lsr_overflow_passes
trap r11
lsr_overflow_passes:
// Overflow set from LSL
ldi $0x05000,r11
ldi $1,r0
rol $30,r0
lsl $1,r0
bv lsl_overflow_passes
trap r11
lsl_overflow_passes:
// Overflow set from LSL, negative to positive
ldi $0x06000,r11
ldi $1,r0
rol $31,r0
lsl $1,r0
bv second_lsl_overflow_passes
trap r11
#endif // OVERFLOW_TEST
#ifdef CARRY_TEST
second_lsl_overflow_passes:
// Test carry
ldi $0x07000,r11
ldi $-1,r0
add $1,r0
tst sys.ccc,cc
trap.z r11
// and carry from subtraction
ldi $0x08000,r11
clr r0
sub $1,r0
tst sys.ccc,cc
trap.z r11
// Carry from right shift
clr r0 ; r0 = 0
lsr 1,r0 ; r0 = 0, c = 0
add.c 1,r0 ; r0 = 0
cmp 1,r0 ; r0 ?= 1
trap.z r11
LDI 1,r0 ; r0 = 1
lsr 1,r0 ; r0 = 0, c = 1
add.c 1,r0 ; r0 = 1
cmp 1,r0
trap.nz r11
ldi 0x070eca6,r0
ldi 0x0408b85,r1
ldi 0x0387653,r2
lsr 1,r0
xor.c r1,r0
cmp r2,r0
trap.nz r11
#endif
#ifdef LOOP_TEST
// Let's try a loop: for i=0; i<5; i++)
// We'll use R0=i, Immediates for 5
ldi $0x09000,r11
clr r0
for_loop:
noop
add $1,r0
cmp $5,r0
blt for_loop
//
// Let's try a reverse loop. Such loops are usually cheaper to
// implement, and this one is no different: 2 loop instructions
// (minus setup instructions) vs 3 from before.
// R0 = 5; (from before)
// do {
// } while (R0 > 0);
ldi $0x0a000,r11
bgt_loop:
noop
sub $1,r0
bgt bgt_loop
// How about the same thing with a >= comparison?
// R1 = 5; // Need to do this explicitly
// do {
// } while(R1 >= 0);
ldi $20,r0
ldi $5,r1
bge_loop:
noop
sub $1,r1
bge bge_loop
// Let's try the reverse loop again, only this time we'll store our
// loop variable in memory.
// R0 = 5; (from before)
// do {
// } while (R0 > 0);
ldi $0x0b000,r11
bra mem_loop_test
loop_var:
.dat 0
mem_loop_test:
mov loop_var(pc),r1
ldi $5,r0
clr r2
sto r0,(r1)
mem_loop:
add $1,r2
add $14,r0
lod (r1),r0
sub $1,r0
sto r0,(r1)
bgt mem_loop
cmp $5,r2
trap.ne r11
#endif
#ifdef SHIFT_TEST
; Now, let's test whether or not our LSR and carry flags work
ldi $0x0c000,r11
ldi -1,r0 ; First test: shifting all the way should yield zero
lsr 32,r0
cmp 0,r0
bnz test_failure
ldi -1,r0 ; Second test: anything greater than zero should set
lsr 0,r0 ; the carry flag
bc test_failure
lsr 1,r0
tst sys.ccc,cc
bz test_failure
lsr 31,r0
tst sys.ccc,cc
bz test_failure
lsr 1,r0
bc test_failure
; Now repeat the above tests, looking to see whether or not ASR works
ldi -1,r0
asr 32,r0
cmp -1,r0
bnz test_failure
ldi -1,r0
asr 0,r0
bc test_failure
cmp -1,r0
bnz test_failure
asr 1,r0
tst sys.ccc,r14
bz test_failure
asr 30,r0
tst sys.ccc,r14
bz test_failure
// Let's test whether LSL works
ldi 0x035,r2
lsl 8,r2
ldi 0x03500,r1
cmp r2,r1
trap.ne r11
ldi 0x074,r0
and 0x0ff,r0
or r0,r2
cmp 0x03574,r2
trap.ne r11
#endif
#ifdef MPY_TEST
// We have two multiply instructions. Let's see if those work
ldi $0x0d000,r11 // Mark our test
ldi 23171,r0 // = sqrt(2)/2 * 32768
mpyu r0,r0 // Should = 2/4 * 2^30 = 2^29 or thereabouts
ldi 536895241,r2
cmp r0,r2
trap.ne r11
ldi 0x0ffff,r0
mpyu r0,r0
ldi 0xfffe0001,r1
cmp r1,r0
trap.ne r11
ldi 0x08001,r0
ldi 0x07fff,r1
mpys r0,r1 // FAILS: result is 0x008001 ??? (pipeline prob)
ldi 0x3fff0001,r2
neg r2
cmp r2,r1 // @0x010011c
trap.ne r11 //TRAP FAILS TO TRIGGER ????? (R2=0x0c000ffff,R1=0x0008001 -- did mpy even happen?)
mpys r0,r0 // FAILS: result is 0x40010001
ldi 0x3fff0001,r2
cmp r2,r0
trap.ne r11 // TRAP FAILS TO TRIGGER AGAIN
ldi 0x08000,r0
mpys r0,r0 // R0 now equals 0x40000000
ldi 0x40000000,r1
cmp r0,r1
trap.ne r11
//
// And from our eyeball test ...
LDI 0x01ff01ff,R0
MOV R0,R7
MOV 8(SP),R6
LSR 7,R0
AND 7,R0
LDI 7,R1
SUB R0,R1
MOV R1,R0
MPYU 5,R0
CMP 20,R0
TRAP.NE R11
#endif
#ifdef PUSH_TEST
ldi $0x0e000,r11 // Mark our test
ldi 0x01248cab,r0
ldi 0xd5312480,r1 // Let's see if we can preserve this as well
mov r1,r7
FJSR(reverse_bit_order,R4); // *SP = 0x010013d
cmp r0,r1
trap.ne r11
cmp r0,r7
trap.ne r11
#endif
#ifdef PIPELINE_STACK_TEST
ldi $0x0f000,r11 // Mark our test
LDI 1,R0
MOV 1(R0),R1
MOV 1(R1),R2
MOV 1(R2),R3
MOV 1(R3),R4
MOV 1(R4),R5
MOV 1(R5),R6
FJSR(pipeline_stack_test,R7)
CMP 1,R0
trap.ne R11
CMP 2,R1
trap.ne R11
CMP 3,R2
trap.ne R11
CMP 4,R3
trap.ne R11
CMP 5,R4
trap.ne R11
CMP 6,R5
trap.ne R11
CMP 7,R6
trap.ne R11
#endif
#ifdef MEM_PIPELINE_TEST
LDI 0x10000,R11
FJSR(mem_pipeline_test,R0)
#endif // MEM_PIPELINE_TEST
#ifdef CONDITIONAL_EXECUTION_TEST
LDI 0x11000,R11
FJSR(conditional_execution_test,R0)
#endif // CONDITIONAL_EXECUTION_TEST
#ifdef NOWAIT_PIPELINE_TEST
LDI 0x12000,R11
FJSR(nowait_pipeline_test,R0)
#endif // NOWAIT_PIPELINE_TEST
#ifdef BCMEM_TEST
LDI 0x13000,R11
CLR R0
LDI -1,R1
STO R0,bcmemtestloc(PC)
LOD bcmemtestloc(PC),R1
CMP R0,R1
TRAP.NZ R11
CMP 0x13000,R11
BZ bcmemtest_cmploc_1
STO R11,bcmemtestloc(PC)
bcmemtest_cmploc_1:
LOD bcmemtestloc(PC),R0
CMP R0,R11
TRAP.Z R11
CLR R0
CMP R0,R11
BZ bcmemtest_cmploc_2
STO.NZ R11,bcmemtestloc(PC)
bcmemtest_cmploc_2:
NOOP
LOD bcmemtestloc(PC),R0
CMP R0,R11
TRAP.NZ R11
BRA end_bcmemtest
bcmemtestloc:
WORD 0
end_bcmemtest:
#endif
// Return success / Test the trap interrupt
clr r11
trap r11 // FAILS HERE FAILS FAILS FAILS !!!!!!!!!!!
noop
noop
busy
// And, in case we miss a halt ...
halt
// Now, let's test whether or not we can handle a subroutine
#ifdef PUSH_TEST
reverse_bit_order:
SUB 3,SP
STO R1,(SP) ; R1 will be our loop counter
STO R2,1(SP) ; R2 will be our accumulator and eventual result
STO R4,2(SP)
LDI 32,R1
CLR R2
reverse_bit_order_loop:
LSL 1,R2
LSR 1,R0
OR.C 1,R2
SUB 1,R1
BNZ reverse_bit_order_loop
MOV R2,R0
LOD (SP),R1
LOD 1(SP),R2
LOD 2(SP),R4
ADD 3,SP
JMP R4
#endif
; The pipeline stack test examines whether or not a series of memory commands
; can be evaluated right after the other without problems. This depends upon
; the calling routine to properly set up registers to be tested.
;
; This is also an incomplete test, as nothing is done to test how these
; pipeline reads/writes are affected by condition codes.
;
#ifdef PIPELINE_STACK_TEST
pipeline_stack_test:
SUB 13,SP
STO R0,(SP)
STO R1,1(SP)
STO R2,2(SP)
STO R3,3(SP)
STO R4,4(SP)
STO R5,5(SP)
STO R6,6(SP)
STO R7,7(SP)
STO R8,8(SP)
STO R9,9(SP)
STO R10,10(SP)
STO R11,11(SP)
STO R12,12(SP)
XOR -1,R0
XOR -1,R1
XOR -1,R2
XOR -1,R3
XOR -1,R4
XOR -1,R5
XOR -1,R6
XOR -1,R7
XOR -1,R8
XOR -1,R9
XOR -1,R10
XOR -1,R11
XOR -1,R12
LOD (SP),R0
LOD 1(SP),R1
LOD 2(SP),R2
LOD 3(SP),R3
LOD 4(SP),R4
LOD 5(SP),R5
LOD 6(SP),R6
LOD 7(SP),R7
LOD 8(SP),R8
LOD 9(SP),R9
LOD 10(SP),R10
LOD 11(SP),R11
LOD 12(SP),R12
ADD 13,SP
JMP R7
#endif // PIPELINE_STACK_TEST
#ifdef MEM_PIPELINE_TEST
mem_pipeline_test:
SUB 4,SP
STO R0,(SP)
STO R1,1(SP)
LDI 0x10000,R11
;
; Test #1 ... Let's start by writing a value to memory
LDI -1,R0
CLR R1
STO R0,2(SP)
LOD 2(SP),R1
CMP R1,R0
MOV.NZ R11,CC
; Test #2, reading and then writing a value from memory
NOP
NOP
CLR R0
CLR R1
LOD 2(SP),R0 ; This should load back up our -1 value
STO R0,3(SP)
; Insist that the pipeline clear
LOD 2(SP),R0
; Now let's try loading into R1
NOP
NOP
NOP
NOP
LOD 3(SP),R1
CMP R1,R0
MOV.NZ R11,CC
LOD (SP),R0
LOD 1(SP),R1
ADD 4,SP
JMP R0
#endif
#ifdef CONDITIONAL_EXECUTION_TEST
conditional_execution_test:
SUB 1,SP
STO R0,(SP)
;
CLRF R0
ADD.Z 1,R0
TRAP.NZ R11
CMP.Z 0,R0
TRAP.Z R11
LOD (SP),R0
ADD 1,SP
JMP R0
#endif
;
; Pipeline stalls have been hideous problems for me. The CPU has been modified
; with special logic to keep stages from stalling. For the most part, this
; means that ALU and memory results may be accessed either before or as they
; are written to the register file. This set of code is designed to test
; whether this bypass logic works.
#ifdef NOWAIT_PIPELINE_TEST
nowait_pipeline_test:
; Allocate for us some number of registers
;
SUB 6,SP
; Leave a spot open on the stack for a local variable,
; kept in memory.
STO R0,(SP)
STO R1,1(SP)
STO R2,2(SP)
STO R3,3(SP)
STO R4,4(SP)
;
; Let's start with ALU-ALU testing
; AA: result->input A
CLR R0
ADD 1,R0
CMP 1,R0
TRAP.NZ R11
; AA: result->input B
CLR R0
CLR R1
ADD 1,R0
CMP R0,R1
TRAP.Z R11
; AA: result->input A on condition
CLRF R0
ADD.Z 5,R0
CMP 5,R0
TRAP.NZ R11
; AA: result->input B on condition
CLR R0
CLRF R1
ADD.Z 5,R0
CMP R0,R1
TRAP.Z R11
; AA: result->input B plus offset
CLR R0
CLRF R1
ADD 5,R0
CMP -5(R0),R1
TRAP.NZ R11
; AA: result->input B plus offset on condition
CLR R0
CLRF R1
ADD.Z 5,R0
CMP -5(R0),R1
TRAP.NZ R11
;
; Then we need to do ALU-Mem input testing
;
CLR R0
STO R0,5(SP)
LDI 8352,R0
LOD 5(SP),R0
TST -1,R0
TRAP.NZ R11
LDI 937,R0 ; Let's try again, this time something that's
STO R0,5(SP) ; not zero
NOOP
LOD 5(SP),R0
CMP 938,R0 ; Let's not compare with self, let's that
TRAP.GE R11 ; masks a problem--compare with a different
CMP 936,R0 ; number instead.
TRAP.LT R11
; Mem output->ALU input testing
; We just did that as partof our last test
; Mem output->MEM input testing
;
LDI 5328,R2
LOD 5(SP),R2
STO R2,5(SP)
LOD 5(SP),R1
CMP 937,R1
TRAP.NZ R11
;
LOD (SP),R0
LOD 1(SP),R1
LOD 2(SP),R2
LOD 3(SP),R3
LOD 4(SP),R4
ADD 6,SP
JMP R0
#endif // NOWAIT_PIPELINE_TEST
fill 512,0
stack: // Must point to a valid word initially
word 0
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