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CS 61C L6.2.2 Interrupts (1) K. Meinz, Summer 2004 © UCB CS61C : Machine Structures Lecture 6.2.2 Interrupts 2004-07-29 Kurt Meinz inst.eecs.berkeley.edu/~cs61c.

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Presentation on theme: "CS 61C L6.2.2 Interrupts (1) K. Meinz, Summer 2004 © UCB CS61C : Machine Structures Lecture 6.2.2 Interrupts 2004-07-29 Kurt Meinz inst.eecs.berkeley.edu/~cs61c."— Presentation transcript:

1 CS 61C L6.2.2 Interrupts (1) K. Meinz, Summer 2004 © UCB CS61C : Machine Structures Lecture 6.2.2 Interrupts 2004-07-29 Kurt Meinz inst.eecs.berkeley.edu/~cs61c

2 CS 61C L6.2.2 Interrupts (2) K. Meinz, Summer 2004 © UCB Big Idea How to choose between associativity, block size, replacement policy? Design against a performance model Minimize: Average Memory Access Time = Hit Time + Miss Penalty x Miss Rate influenced by technology & program behavior Note: Hit Time encompasses Hit Rate!!! Create the illusion of a memory that is large, cheap, and fast - on average

3 CS 61C L6.2.2 Interrupts (3) K. Meinz, Summer 2004 © UCB Example Assume Hit Time = 1 cycle Miss rate = 5% Miss penalty = 20 cycles (on top of hit) Calculate AMAT… Avg mem access time = 1 + 0.05 x 20 = 1 + 1 cycles = 2 cycles

4 CS 61C L6.2.2 Interrupts (4) K. Meinz, Summer 2004 © UCB Ways to reduce miss rate Larger cache limited by cost and technology hit time of first level cache < cycle time More places in the cache to put each block of memory – associativity fully-associative -any block any line k-way set associated -k places for each block -direct map: k=1

5 CS 61C L6.2.2 Interrupts (5) K. Meinz, Summer 2004 © UCB Improving Miss Penalty When caches first became popular, Miss Penalty ~ 10 processor clock cycles Today 2400 MHz Processor (0.4 ns per clock cycle) and 80 ns to go to DRAM  200 processor clock cycles! Proc $2$2 DRAM $ MEM Solution: another cache between memory and the processor cache: Second Level (L2) Cache

6 CS 61C L6.2.2 Interrupts (6) K. Meinz, Summer 2004 © UCB Analyzing Multi-level cache hierarchy Proc $2$2 DRAM $ L1 hit time L1 Miss Rate L1 Miss Penalty Avg Mem Access Time = L1 Hit Time + L1 Miss Rate * L1 Miss Penalty L1 Miss Penalty = AMAT L2 = L2 Hit Time + L2 Miss Rate * L2 Miss Penalty Avg Mem Access Time = L1 Hit Time + L1 Miss Rate * (L2 Hit Time + L2 Miss Rate * L2 Miss Penalty ) L2 hit time L2 Miss Rate L2 Miss Penalty

7 CS 61C L6.2.2 Interrupts (7) K. Meinz, Summer 2004 © UCB Typical Scale L1 size: tens of KB hit time: complete in one clock cycle miss rates: 1-5% L2: size: hundreds of KB hit time: few clock cycles miss rates: 10-20% L2 miss rate is fraction of L1 misses that also miss in L2 why so high?

8 CS 61C L6.2.2 Interrupts (8) K. Meinz, Summer 2004 © UCB Example: with L2 cache Assume L1 Hit Time = 1 cycle L1 Miss rate = 5% L2 Hit Time = 5 cycles L2 Miss rate = 15% (% L1 misses that miss) L2 Miss Penalty = 200 cycles L1 miss penalty = 5 + 0.15 * 200 = 35 Avg mem access time = 1 + 0.05 x 35 = 2.75 cycles

9 CS 61C L6.2.2 Interrupts (9) K. Meinz, Summer 2004 © UCB Example: without L2 cache Assume L1 Hit Time = 1 cycle L1 Miss rate = 5% L1 Miss Penalty = 200 cycles Avg mem access time = 1 + 0.05 x 200 = 11 cycles 4x faster with L2 cache! (2.75 vs. 11)

10 CS 61C L6.2.2 Interrupts (10) K. Meinz, Summer 2004 © UCB Cache Summary Cache design choices: size of cache: speed v. capacity direct-mapped v. associative for N-way set assoc: choice of N block replacement policy 2nd level cache? Write through v. write back? Use performance model to pick between choices, depending on programs, technology, budget,...

11 CS 61C L6.2.2 Interrupts (11) K. Meinz, Summer 2004 © UCB Outline Exceptions Memory Mapped IO Exception Implementation

12 CS 61C L6.2.2 Interrupts (12) K. Meinz, Summer 2004 © UCB Recall : 5 components of any Computer Processor (active) Computer Control (“brain”) Datapath (“brawn”) Memory (passive) (where programs, data live when running) Devices Input Output Keyboard, Mouse Display, Printer Disk, Network

13 CS 61C L6.2.2 Interrupts (13) K. Meinz, Summer 2004 © UCB Motivation for Input/Output I/O is how humans interact with computers I/O gives computers long-term memory. I/O lets computers do amazing things: Read pressure of synthetic hand and control synthetic arm and hand of fireman Control propellers, fins, communicate in BOB (Breathable Observable Bubble) Computer without I/O like a car without wheels; great technology, but won’t get you anywhere

14 CS 61C L6.2.2 Interrupts (14) K. Meinz, Summer 2004 © UCB I/O Device Examples and Speeds I/O Speed: bytes transferred per second (from mouse to Gigabit LAN: 100-million-to-1) DeviceBehaviorPartner Data Rate (KBytes/s) KeyboardInputHuman0.01 MouseInputHuman0.02 Voice outputOutputHuman5.00 Floppy diskStorageMachine50.00 Laser PrinterOutputHuman100.00 Magnetic DiskStorageMachine10,000.00 Wireless NetworkI or OMachine 10,000.00 Graphics DisplayOutputHuman30,000.00 Wired LAN NetworkI or OMachine1,000,000.00

15 CS 61C L6.2.2 Interrupts (15) K. Meinz, Summer 2004 © UCB What do we need to make I/O work? A way to present them to user programs so they are useful cmd reg. data reg. Operating System APIsFiles Proc Mem A way to connect many types of devices to the Proc-Mem PCI Bus SCSI Bus A way to control these devices, respond to them, and transfer data

16 CS 61C L6.2.2 Interrupts (16) K. Meinz, Summer 2004 © UCB Instruction Set Architecture for I/O What must the processor do for I/O? Input: reads a sequence of bytes Output: writes a sequence of bytes Some processors have special input and output instructions Alternative model (used by MIPS): Use loads for input, stores for output Called “Memory Mapped Input/Output” A portion of the address space dedicated to communication paths to Input or Output devices (no memory there)

17 CS 61C L6.2.2 Interrupts (17) K. Meinz, Summer 2004 © UCB Memory Mapped I/O Certain addresses are not regular memory Instead, they correspond to registers in I/O devices cntrl reg. data reg. 0 0xFFFFFFFF 0xFFFF0000 address

18 CS 61C L6.2.2 Interrupts (18) K. Meinz, Summer 2004 © UCB Processor-I/O Speed Mismatch 1GHz microprocessor can execute 1 billion load or store instructions per second, or 4,000,000 KB/s data rate I/O devices data rates range from 0.01 KB/s to 1,000,000 KB/s Input: device may not be ready to send data as fast as the processor loads it Also, might be waiting for human to act Output: device not be ready to accept data as fast as processor stores it What to do?

19 CS 61C L6.2.2 Interrupts (19) K. Meinz, Summer 2004 © UCB Processor Checks Status before Acting Path to device generally has 2 registers: Control Register, says it’s OK to read/write (I/O ready) [think of a flagman on a road] Data Register, contains data Processor reads from Control Register in loop, waiting for device to set Ready bit in Control reg (0  1) to say its OK Processor then loads from (input) or writes to (output) data register Load from or Store into Data Register  Proc resets Ready bit (1  0) of Control Register

20 CS 61C L6.2.2 Interrupts (20) K. Meinz, Summer 2004 © UCB SPIM/Proj4 I/O Simulation Simulate 1 I/O device: memory-mapped terminal (keyboard + display) Read from keyboard (receiver); 2 device regs Writes to terminal (transmitter); 2 device regs Received Byte Receiver Data 0xffff0004 Unused (00...00) (IE) Receiver Control 0xffff0000 Ready (I.E.) Unused (00...00) Transmitted Byte Transmitter Control 0xffff0008 Transmitter Data 0xffff000c Ready (I.E.) Unused (00...00) Unused

21 CS 61C L6.2.2 Interrupts (21) K. Meinz, Summer 2004 © UCB SPIM I/O Control register rightmost bit (0): Ready Receiver: Ready==1 means character in Data Register not yet been read; 1  0 when data is read from Data Reg Transmitter: Ready==1 means transmitter is ready to accept a new character; 0  Transmitter still busy writing last char -I.E. bit discussed later Data register rightmost byte has data Receiver: last char from keyboard; rest = 0 Transmitter: when write rightmost byte, writes char to display

22 CS 61C L6.2.2 Interrupts (22) K. Meinz, Summer 2004 © UCB I/O Example Input: Read from keyboard into $v0 lui$t0, 0xffff #ffff0000 Waitloop:lw$t1, 0($t0) #control andi$t1,$t1,0x1 beq$t1,$zero, Waitloop lw$v0, 4($t0) #data Output: Write to display from $a0 lui$t0, 0xffff #ffff0000 Waitloop:lw$t1, 8($t0) #control andi$t1,$t1,0x1 beq$t1,$zero, Waitloop sw$a0, 12($t0) #data Processor waiting for I/O called “Polling”

23 CS 61C L6.2.2 Interrupts (23) K. Meinz, Summer 2004 © UCB Cost of Polling? Assume for a processor with a 1GHz clock it takes 400 clock cycles for a polling operation (call polling routine, accessing the device, and returning). Determine % of processor time for polling Mouse: polled 30 times/sec so as not to miss user movement Floppy disk: transfers data in 2-Byte units and has a data rate of 50 KB/second. No data transfer can be missed. Hard disk: transfers data in 16-Byte chunks and can transfer at 16 MB/second. Again, no transfer can be missed.

24 CS 61C L6.2.2 Interrupts (24) K. Meinz, Summer 2004 © UCB % Processor time to poll [p. 677 in book] Mouse Polling, Clocks/sec = 30 [polls/s] * 400 [clocks/poll] = 12K [clocks/s] % Processor for polling: 12*10 3 [clocks/s] / 1*10 9 [clocks/s] = 0.0012%  Polling mouse little impact on processor Frequency of Polling Floppy = 50 [KB/s] / 2 [B/poll] = 25K [polls/s] Floppy Polling, Clocks/sec = 25K [polls/s] * 400 [clocks/poll] = 10M [clocks/s] % Processor for polling: 10*10 6 [clocks/s] / 1*10 9 [clocks/s] = 1%  OK if not too many I/O devices

25 CS 61C L6.2.2 Interrupts (25) K. Meinz, Summer 2004 © UCB % Processor time to poll hard disk Frequency of Polling Disk = 16 [MB/s] / 16 [B] = 1M [polls/s] Disk Polling, Clocks/sec = 1M [polls/s] * 400 [clocks/poll] = 400M [clocks/s] % Processor for polling: 400*10 6 [clocks/s] / 1*10 9 [clocks/s] = 40%  Unacceptable

26 CS 61C L6.2.2 Interrupts (26) K. Meinz, Summer 2004 © UCB What is the alternative to polling? Wasteful to have processor spend most of its time “spin-waiting” for I/O to be ready Would like an unplanned procedure call that would be invoked only when I/O device is ready Solution: use exception mechanism to help I/O. Interrupt program when I/O ready, return when done with data transfer

27 CS 61C L6.2.2 Interrupts (27) K. Meinz, Summer 2004 © UCB Definitions for Clarification Exception: signal marking that something “out of the ordinary” has happened and needs to be handled Interrupt: asynchronous exception Trap: synchronous exception Note: These are different from the book’s definitions. All I care about: that you know the difference between sync and async.

28 CS 61C L6.2.2 Interrupts (28) K. Meinz, Summer 2004 © UCB I/O Interrupt An I/O interrupt is like overflow exceptions except: An I/O interrupt is asynchronous More information needs to be conveyed An I/O interrupt is asynchronous with respect to instruction execution: I/O interrupt is not associated with any instruction, but it can happen in the middle of any given instruction I/O interrupt does not prevent any instruction from completion

29 CS 61C L6.2.2 Interrupts (29) K. Meinz, Summer 2004 © UCB Interrupt Driven Data Transfer (1) I/O interrupt (2) save PC Memory add sub and or user program read store... jr interrupt service routine (3) jump to interrupt service routine (4) perform transfer (5)

30 CS 61C L6.2.2 Interrupts (30) K. Meinz, Summer 2004 © UCB SPIM I/O Simulation: Interrupt Driven I/O I.E. stands for Interrupt Enable Set Interrupt Enable bit to 1 have interrupt occur whenever Ready bit is set Received Byte Receiver Data 0xffff0004 Unused (00...00) (IE) Receiver Control 0xffff0000 Ready (I.E.) Unused (00...00) Transmitted Byte Transmitter Control 0xffff0008 Transmitter Data 0xffff000c Ready (I.E.) Unused (00...00) Unused

31 CS 61C L6.2.2 Interrupts (31) K. Meinz, Summer 2004 © UCB Benefit of Interrupt-Driven I/O Find the % of processor consumed if the hard disk is only active 5% of the time. Assuming 500 clock cycle overhead for each transfer, including interrupt: Disk Interrupts/s = 16 MB/s / 16B/interrupt = 1M interrupts/s Disk Interrupts, clocks/s = 1M interrupts/s * 500 clocks/interrupt = 500,000,000 clocks/s % Processor for during transfer: 500*10 6 / 1*10 9 = 50% Disk active 5%  5% * 50%  2.5% busy

32 CS 61C L6.2.2 Interrupts (32) K. Meinz, Summer 2004 © UCB Generalizing Interrupts We can handle all sorts of exceptions with interrupts. Big idea: jump to handler that knows what to do with each interrupt, then jump back Our types: syscall, overflow, mmio ready.

33 CS 61C L6.2.2 Interrupts (33) K. Meinz, Summer 2004 © UCB OS: I/O Requirements The OS must be able to prevent: The user program from communicating with the I/O device directly If user programs could perform I/O directly: No protection to the shared I/O resources 3 types of communication are required: The OS must be able to give commands to the I/O devices The I/O device notify OS when the I/O device has completed an operation or an error Data transfers between memory and I/O device

34 CS 61C L6.2.2 Interrupts (34) K. Meinz, Summer 2004 © UCB Instruction Set Support for OS (1/2) How to turn off interrupts during interrupt routine? Bit in Status Register determines whether or not interrupts enabled: Interrupt Enable bit (IE) (0  off, 1  on) Status Register(described later) IE

35 CS 61C L6.2.2 Interrupts (35) K. Meinz, Summer 2004 © UCB Instruction Set Support for OS (2/2) How to prevent user program from turning off interrupts (forever)? Bit in Status Register determines whether in user mode or OS (kernel) mode: Kernel/User bit (KU) (0  kernel, 1  user) Status RegisterAssume Unused IEKU On exception/interrupt disable interrupts (IE=0) and go into kernel mode (KU=0)

36 CS 61C L6.2.2 Interrupts (36) K. Meinz, Summer 2004 © UCB Kernel/User Mode Generally restrict device access to OS HOW? Add a “mode bit” to the machine: K/U Only allow SW in “kernel mode” to access device registers If user programs could access device directly? could destroy each others data,... might break the devices, …

37 CS 61C L6.2.2 Interrupts (37) K. Meinz, Summer 2004 © UCB Crossing the System Boundary System loads user program into memory and ‘gives’ it use of the processor Switch back SYSCALL -request service -I/O TRAP (overflow) Interrupt Proc Mem I/O Bus cmd reg. data reg. System User

38 CS 61C L6.2.2 Interrupts (38) K. Meinz, Summer 2004 © UCB Syscall How does user invoke the OS? syscall instruction: invoke the kernel (Go to 0x80000080, change to kernel mode) By software convention, $v0 has system service requested: OS performs request

39 CS 61C L6.2.2 Interrupts (39) K. Meinz, Summer 2004 © UCB SPIM OS Services via Syscall Note: most OS services deal with I/O print_int1 $a0 = integer print_float2 $f12 = float print_double3 $f12 = double print_string4 $a0 = string read_int5integer (in $v0 ) read_float6float (in $f0 ) read_double7double (in $f0 ) read_string8 $a0 = buffer, $a1 = length sbrk9 $a0 = amountaddress(in $v0 ) exit10 ServiceCodeArgsResult (put in $v0 )

40 CS 61C L6.2.2 Interrupts (40) K. Meinz, Summer 2004 © UCB Example: User invokes OS (SPIM) Print “the answer = 42” First print “the answer =”:.data str:.asciiz"the answer = ".text li $v0,4# 4=code for print_str la$a0,str# address of string syscall# print the string Now print 42 li$v0,1# 1=code for print_int li$a0,42# integer to print syscall# print int

41 CS 61C L6.2.2 Interrupts (41) K. Meinz, Summer 2004 © UCB Handling a Single Interrupt (1/3) An interrupt has occurred, then what? Automatically, the hardware copies PC into EPC ($14 on cop0) and puts correct code into Cause Reg ($13 on cop0) Automatically, PC is set to 0x80000080, process enters kernel mode, and interrupt handler code begins execution Interrupt Handler code: Checks Cause Register (bits 5 to 2 of $13 in cop0) and jumps to portion of interrupt handler which handles the current exception

42 CS 61C L6.2.2 Interrupts (42) K. Meinz, Summer 2004 © UCB Handling a Single Interrupt (2/3) Sample Interrupt Handler Code.text 0x80000080 mfc0 $k0,$13 # $13 is Cause Reg sll $k0,$k0,26 # isolate srl $k0,$k0,28 # Cause bits Notes: Don’t need to save $k0 or $k1 -MIPS software convention to provide temp registers for operating system routines -Application software cannot use them Can only work on CPU, not on cop0

43 CS 61C L6.2.2 Interrupts (43) K. Meinz, Summer 2004 © UCB Handling a Single Interrupt (3/3) When the interrupt is handled, copy the value from EPC to the PC. Call instruction rfe (return from exception), which will return process to user mode and reset state to the way it was before the interrupt What about multiple interrupts?

44 CS 61C L6.2.2 Interrupts (44) K. Meinz, Summer 2004 © UCB Multiple Interrupts Problem: What if we’re handling an Overflow interrupt and an I/O interrupt (printer ready, for example) comes in? Options: drop any conflicting interrupts: unrealistic, they may be important simultaneously handle multiple interrupts: unrealistic, may not be able to synchronize them (such as with multiple I/O interrupts) queue them for later handling: sounds good

45 CS 61C L6.2.2 Interrupts (45) K. Meinz, Summer 2004 © UCB Prioritized Interrupts (1/3) Question: Suppose we’re dealing with a computer running a nuclear facility. What if we’re handling an Overflow interrupt and a Nuclear Meltdown Imminent interrupt comes in? Answer: We need to categorize and prioritize interrupts so we can handle them in order of urgency: emergency vs. luxury.

46 CS 61C L6.2.2 Interrupts (46) K. Meinz, Summer 2004 © UCB Prioritized Interrupts (2/3) OS convention to simplify software: Process cannot be preempted by interrupt at same or lower "level" Return to interrupted code as soon as no more interrupts at a higher level When an interrupt is handled, take the highest priority interrupt on the queue -may be partially handled, may not, so we may need to save state of interrupts(!)

47 CS 61C L6.2.2 Interrupts (47) K. Meinz, Summer 2004 © UCB Prioritized Interrupts (3/3) To implement, we need an Exception Stack: portion of address space allocated for stack of “Exception Frames” each frame represents one interrupt: contains priority level as well as enough info to restart handling it if necessary

48 CS 61C L6.2.2 Interrupts (48) K. Meinz, Summer 2004 © UCB Modified Interrupt Handler (1/3) Problem: When an interrupt comes in, EPC and Cause get overwritten immediately by hardware. Lost EPC means loss of user program. Solution: Modify interrupt handler. When first interrupt comes in: disable interrupts (in Status Register) save EPC, Cause, Status and Priority Level on Exception Stack re-enable interrupts continue handling current interrupt

49 CS 61C L6.2.2 Interrupts (49) K. Meinz, Summer 2004 © UCB Modified Interrupt Handler (2/3) When next (or any later) interrupt comes in: interrupt the first one disable interrupts (in Status Register) save EPC, Cause, Status and Priority Level (and maybe more) on Exception Stack determine whether new one preempts old one -if no, re-enable interrupts and continue with old one -if yes, may have to save state for the old one, then re-enable interrupts, then handle new one

50 CS 61C L6.2.2 Interrupts (50) K. Meinz, Summer 2004 © UCB Modified Interrupt Handler (3/3) Notes: Disabling interrupts is dangerous So we disable them for as short a time as possible: long enough to save vital info onto Exception Stack This new scheme allows us to handle many interrupts effectively.

51 CS 61C L6.2.2 Interrupts (51) K. Meinz, Summer 2004 © UCB Interrupt Levels in MIPS? What are they? It depends what the MIPS chip is inside of: differ by app Casio PalmPC, Sony Playstation, HP LaserJet printer MIPS architecture enables priorities for different I/O events

52 CS 61C L6.2.2 Interrupts (52) K. Meinz, Summer 2004 © UCB Interrupt Levels in MIPS Architecture Conventionally, from highest level to lowest level exception/interrupt levels: Bus error Illegal Instruction/Address trap High priority I/O Interrupt (fast response) Low priority I/O Interrupt (slow response) (later in course, will talk about other events with other levels)

53 CS 61C L6.2.2 Interrupts (53) K. Meinz, Summer 2004 © UCB Improving Data Transfer Performance Thus far: OS give commands to I/O, I/O device notify OS when the I/O device completed operation or an error What about data transfer to I/O device? Processor busy doing loads/stores between memory and I/O Data Register Ideal: specify the block of memory to be transferred, be notified on completion? Direct Memory Access (DMA) : a simple computer transfers a block of data to/from memory and I/O, interrupting upon done

54 CS 61C L6.2.2 Interrupts (54) K. Meinz, Summer 2004 © UCB Example: code in DMA controller DMA code from Disk Device to Memory.data Count:.word 4096 Start:.space 4096 DMA “computer” in parallel with CPU.text Initial:lw $s0, Count# No. chars la $s1, Start# @next char Wait:lw $s2, DiskControl andi $s2,$s2,1 # select Ready beq $s2,$0,Wait # spinwait lb $t0, DiskData # get byte sb $t0, 0($s1) # transfer addiu $s0,$s0,-1 # Count-- addiu $s1,$s1,1 # Start++ bne $s0,$0,Wait # next char

55 CS 61C L6.2.2 Interrupts (55) K. Meinz, Summer 2004 © UCB Details not covered MIPS has a field to record all pending interrupts so that none are lost while interrupts are off; in Cause register The Interrupt Priority Level that the CPU is running at is set in memory MIPS has a field in that can mask interrupts of different priorities to implement priority levels; in Status register MIPS has limited nesting of saving KU,IE bits to recall in case higher priority interrupts; in Status Register

56 CS 61C L6.2.2 Interrupts (56) K. Meinz, Summer 2004 © UCB Interrupts while serving interrupts? Suppose there was an interrupt while the interrupt enable or mask bit is off: what should you do? (cannot ignore) Cause register has field--Pending Interrupts (PI)-- 5 bits wide (bits15:11) for each of the 5 HW interrupt levels Bit becomes 1 when an interrupt at its level has occurred but not yet serviced Interrupt routine checks pending interrupts ANDed with interrupt mask to decide what to service Cause Register ExcCodePI

57 CS 61C L6.2.2 Interrupts (57) K. Meinz, Summer 2004 © UCB support for OS: User => System mode Bit in Status Register determines whether in user mode or OS (kernel) mode: Kernel/User bit (KU) (0  kernel, 1  user) Status Register(described later) IEKU On exception/interrupt disable interrupts (IE=0) and go into kernel mode (KU=0) How remember old KU, IE bits? Hardware copies Current KU and IE bits (0-1) into Previous KU and IE bits (2-3) Status Register (described later) IEKUIEKU 00

58 CS 61C L6.2.2 Interrupts (58) K. Meinz, Summer 2004 © UCB support for OS: System => user mode OS saves user registers, performs its task and restores user registers. can JR back to value saved in EPC how to get back to user mode? use Return from Exception ( rfe ) Status Register (described later) IE KUKU KUKU

59 CS 61C L6.2.2 Interrupts (59) K. Meinz, Summer 2004 © UCB Prioritizing Interrupts How implement interrupt levels? Allow selective interruption via Interrupt Mask(IM) in Status Register: 5 for HW interrupts Interrupt only if IE==1 AND Mask bit == 1 (bits 15:11 of SR) for that interrupt level Set Mask Bits above your level to 1 To support interrupts of interrupts, have 3 deep stack in Status for IE,K/U bits: Current (1:0), Previous (3:2), Old (5:4) IEKU Status Register IEKUIEKU IM CPO 00

60 CS 61C L6.2.2 Interrupts (60) K. Meinz, Summer 2004 © UCB Revised Interrupt Routine 2/2 Jump to appropriate interrupt routine On Return, disable interrupts using Current IE bit of Status Register Then restore saved registers, previous KU,IE bits of Status (via rfe ) and return to instruction determined by old EPC rfe IEKUIEKUIEKU IM Status Reg After IEKU Status Reg Before IEKUIEKU IM CurrentPre.Old

61 CS 61C L6.2.2 Interrupts (61) K. Meinz, Summer 2004 © UCB Re-entrant Interrupt Routine? How allow interrupt of interrupts and safely save registers? Stack? Resources consumed by each exception, so cannot tolerate arbitrary deep nesting of exceptions/interrupts With priority level system only interrupted by higher priority interrupt, so cannot be recursive  Only need one save area (“exception frame”) per priority level

62 CS 61C L6.2.2 Interrupts (62) K. Meinz, Summer 2004 © UCB Things to Remember Kernel Mode v. User Mode: OS can provide security and fairness Syscall: provides a way for a programmer to avoid having to know details of each I/O device To be acceptable, interrupt handler must: service all interrupts (no drops) service by priority make all users believe that no interrupt has occurred

63 CS 61C L6.2.2 Interrupts (63) K. Meinz, Summer 2004 © UCB Interrupts Interrupt Implementation

64 CS 61C L6.2.2 Interrupts (64) K. Meinz, Summer 2004 © UCB Generalizing Interrupts Must support: Jumping to handler -On exception, proc sets nPC  handlerAddr Knowing what happened/bookkeeping -EPC: Reg holds ~inst at which nrpt occurred -Cause: Holds the specific interrupt Jumping back to user program -More bookkeeping (back to user mode) - ~ “jr $EPC”

65 CS 61C L6.2.2 Interrupts (65) K. Meinz, Summer 2004 © UCB Knowing what happened On exception, proc copies PC+4 into EPC: assign exception = arith_excp | mem_excp | sys_excp always @ (posedge clk) if (exception) EPC <= PC + 4; Proc copies exception number into cause: always @ (posedge clk) { if (arith_excp) Cause <= 0x1; else if (mem_excp) Cause <= 0x2; else if (sys_excp) Cause <= 0x4; }

66 CS 61C L6.2.2 Interrupts (66) K. Meinz, Summer 2004 © UCB Jumping to handler New nPC mux: assign nPC = (exception) ? 0x04000040 : old_npc; Overrides all other nPC signals Jump to handler takes priority over branch.

67 CS 61C L6.2.2 Interrupts (67) K. Meinz, Summer 2004 © UCB In handler: Handler uses special regs/instrs to access exception data: $k0/$k1 (Why not $t0 … $t9?) mfc0 $reg $EPC mfc0 $reg $CAUSE When done, move EPC (or whatever) into $k0 and then jr $k0.

68 CS 61C L6.2.2 Interrupts (68) K. Meinz, Summer 2004 © UCB Proc Support for mfc0: mfc0 $reg X Lots of options: In decode as “add $reg X $0”: -Mux in front of BusA <- ForwardA, Cause, EPC -Mux in front of BusB <- ForwardB, 0 In wb: -Mux in front of regdst: Regdst <- ALUout, MemOut, Cause, EPC  Which one do you think pipelines better?

69 CS 61C L6.2.2 Interrupts (69) K. Meinz, Summer 2004 © UCB What about pipelining?! Precise Exceptions  State of the machine is preserved as if program executed up to (not including) the offending instruction All previous instructions completed Offending instruction and all following instructions act as if they have not even started Same system code will work on different implementations

70 CS 61C L6.2.2 Interrupts (70) K. Meinz, Summer 2004 © UCB Exception/Interrupts: Implementation 5 instructions, executing in 5 different pipeline stages! Who caused the interrupt? StageProblem interrupts occurring IFPage fault on instruction fetch; misaligned memory access; memory-protection violation IDUndefined or illegal opcode; syscall EXArithmetic exception MEMPage fault on data fetch; misaligned memory access; memory-protection violation; memory error How do we stop the pipeline? How do we restart it? Do we interrupt immediately or wait? How do we sort all of this out to maintain preciseness?

71 CS 61C L6.2.2 Interrupts (71) K. Meinz, Summer 2004 © UCB Exception Handling npc I mem Regs B alu S D mem m IAU PC lw $2,20($5) Regs A imoprwn detect bad instruction address detect bad instruction detect overflow detect bad data address Allow exception to take effect Excp

72 CS 61C L6.2.2 Interrupts (72) K. Meinz, Summer 2004 © UCB Another look at the exception problem Use pipeline to sort this out! Pass exception status along with instruction. Keep track of PCs for every instruction in pipeline. Don’t act on exception until it reaches WB stage When instruction reaches WB stage: Save PC  EPC, exc_status  cause, handler addr  PC Turn all instructions in earlier stages into noops! Program Flow Time IFetchDcdExecMemWB IFetchDcdExecMemWB IFetchDcdExecMemWB IFetchDcdExecMemWB Data TLB Bad Inst Inst TLB fault Overflow

73 CS 61C L6.2.2 Interrupts (73) K. Meinz, Summer 2004 © UCB Detailed implementation New pipeline regs: Valid <- If instr. Is not valid, no writes to regfile or mem will occur. Invalids will not trigger exceptions iPC <- addr of the current instr. Will go into EPC Exception <- type of exception. Will go into cause. In each stage: if excepting, set except reg. “flag exception” In MEM, WB, only do write if Valid In WB: “raise exception” …

74 CS 61C L6.2.2 Interrupts (74) K. Meinz, Summer 2004 © UCB Detailed implementation In WB: if (ME/WB.Valid & ME/WB.Except) { IF/DE.valid_in = DE/Ex.Valid_in = EX/ME.Valid_in = ME/WB.valid_in = 0; npc  ME/WB.iPC; cause  Me/WB.Except; EPC  Me/WB.iPC;


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