Quiz 0 Lecture 2: The Instruction Set Architecture COS / ELE 375 Computer Architecture and Organization Princeton University Fall 2015 Prof. David August 1 2 Quiz 0 CD 3 Miles of Music 3 4 Pits and Lands Interpretation 0 1 1 1 0 1 0 1 As Music: 011101012 = 117/256 position of speaker As Number: Transition represents a bit state (1/on/red/female/heads) 01110101 = 1 + 4 + 16 + 32 + 64 = 117 = 75 No change represents other state (0/off/white/male/tails) 2 10 16 (Get comfortable with base 2, 8, 10, and 16.) As Text: th 011101012 = 117 character in the ASCII codes = “u” 5 6 Interpretation – ASCII Princeton Computer Science Building West Wall 7 8 Interpretation Binary Code and Data (Hello World!) • Programs consist of Code and Data • Code and Data are Encoded in Bits IA-64 Binary (objdump) As Music: 011101012 = 117/256 position of speaker As Number: 011101012 = 1 + 4 + 16 + 32 + 64 = 11710 = 7516 As Text: th 011101012 = 117 character in the ASCII codes = “u” CAN ALSO BE INTERPRETED AS MACHINE INSTRUCTION! 9 Interfaces in Computer Systems Instructions Sequential Circuit!! Software: Produce Bits Instructing Machine to Manipulate State or Produce I/O Computers process information State Applications • Input/Output (I/O) Operating System • State (memory) • Computation (processor) Compiler Firmware Instruction Set Architecture Input Output Instruction Set Processor I/O System Datapath & Control Computation Digital Design Circuit Design • Instructions instruct processor to manipulate state Layout • Instructions instruct processor to produce I/O in the same way Hardware: Read and Obey Instruction Bits 12 State State – Main Memory Typical modern machine has this architectural state: Main Memory (AKA: RAM – Random Access Memory) 1. Main Memory • Data can be accessed by address (like a big array) 2. Registers • Large but relatively slow 3. Program Counter • Decent desktop machine: 1 Gigabyte, 800MHz Address Data 0000 01011001 Architectural – Part of the assembly programmer’s interface 2 0001 F5 (Implementation has additional microarchitectural state) 16 0002 7816 0003 3A16 … … FFFF 000000002 Byte Addressable 13 14 State – Main Memory State – Main Memory Address Data Address Data Read: Write: 0000 01011001 0000 01011001 1. Indicate READ 2 1. Indicate WRITE 2 0001 F5 0001 F5 2. Give Address 16 2. Give Address and Data 16 0002 78 0002 78 3. Get Data 16 16 0003 3A16 0003 3A121616 … … … … FFFF 000000002 FFFF 000000002 Read/Write Read/Write READ WRITE Address Address 0002 0003 Data Data 7816 1216 15 16 State – Registers State – Program Counter Registers (AKA: Register File) • Data can be accessed by register number (address) Program Counter (AKA: PC, Instruction Pointer, IP) • Small but relatively fast (typically on processor chip) • Instructions change state, but which instruction now? • Decent desktop machine: 8 32-bit registers, 3 GHz • PC holds memory address of currently executing instruction Register Data in Reg Address Data in Memory 0 0000000016 Program Counter 0000 010110012 1 F629D9B516 0002 0001 F516 2 7B2D9D0816 0002 ADDinst 3 0000000116 0003 SUBTRACTinst … … … … 8 DEADBEEF16 FFFF 000000002 17 18 State – Program Counter State – Summary Program Counter (AKA: PC, Instruction Pointer, IP) Typical modern machine has this architectural state: • Instructions change state, but which instruction now? 1. Main Memory – Big, Slow • PC holds address of currently executing instruction 2. Registers – Small, Fast (always on processor chip) • PC is updated after each instruction 3. Program Counter – Address of executing instruction Address Data in Memory Architectural – Part of the assembly programmer’s Program Counter 0000 010110012 interface 0003 0001 F516 (implementation has additional microarchitectural state) 0002 ADDinst 0003 SUBTRACTinst … … FFFF 000000002 19 20 An Aside: State and The Core Dump Interfaces in Computer Systems Software: Produce Bits Instructing Machine to Manipulate State or Produce I/O • Core Dump: the state of the machine at a given time Applications Registers • Typically at program failure Operating System 0 1 2 3 4 5 6 7 • Core dump contains: 0000 0788 B700 0010 0401 0002 0003 00A0 • Register Contents PC 8 9 A B C D E F Compiler Firmware 10 0000 0788 B700 0010 0401 0002 0003 00A0 • Memory Contents Instruction Set Architecture • PC Value Main Memory Instruction Set Processor I/O System 00: 0000 0000 0000 0000 0000 0000 0000 0000 Datapath & Control 08: 0000 0000 0000 0000 0000 0000 0000 0000 10: 9222 9120 1121 A120 1121 A121 7211 0000 Digital Design 18: 0000 0001 0002 0003 0004 0005 0006 0007 20: 0008 0009 000A 000B 000C 000D 000E 000F Circuit Design 28: 0000 0000 0000 FE10 FACE CAFE ACED CEDE Layout . E8: 1234 5678 9ABC DEF0 0000 0000 F00D 0000 Hardware: Read and Obey Instruction Bits F0: 0000 0000 EEEE 1111 EEEE 1111 0000 0000 F8: B1B2 F1F5 0000 0000 0000 0000 0000 0000 21 Instructions Instructions Register Data An ADD Instruction: Instructions: 0 0 add r1 = r2 + r3 (assembly) “The vocabulary of commands” 1 15 Specify how to operate on state 2 1 Opcode Operands 3 2 Example: … … 40: add r1 = r2 + r3 31 0 Parts of the Instruction: 44: sub r3 = r1 - r0 Address Data • Opcode (verb) – what operation to perform 48: store M[ r3 ] = r1 0 0 • Operands (noun) – what to operate upon 52: load r2 = M[ 2 ] 1 25 • Source Operands – where values come from 2 5 • Destination Operand – where to deposit data values Program Counter 3 9 40 … … FFFFFFFF 0 Instructions Instructions Register Data Register Data Instructions: 0 0 0 0 “The vocabulary of commands” 1 15 1 3 Specify how to operate on state 2 1 2 1 3 2 3 2 Example: … … Example: … … 31 0 40: add r13 = r2 + r3 40: add r1 = r2 + r3 31 0 44: sub r3 = r1 - r0 Address Data 44: sub r3 = r1 - r0 Address Data 48: store M[ r3 ] = r1 48: store M[ r3 ] = r1 0 0 0 0 52: load r2 = M[ 2 ] 52: load r2 = M[ 2 ] 1 25 1 25 2 5 2 5 Program Counter 3 9 Program Counter 3 9 40 … … 40 … … FFFFFFFF 0 FFFFFFFF 0 Instructions Instructions Register Data Register Data Instructions: 0 0 Instructions: 0 0 “The vocabulary of commands” 1 3 “The vocabulary of commands” 1 3 Specify how to operate on state 2 1 Specify how to operate on state 2 1 3 2 3 3 Example: … … Example: … … 40: add r1 = r2 + r3 31 0 40: add r1 = r2 + r3 31 0 44: sub r33 = r1 - r0 Address Data 44: sub r3 = r1 - r0 Address Data 48: store M[ r3 ] = r1 48: store M[ r3 ] = r13 0 0 0 0 52: load r2 = M[ 2 ] 52: load r2 = M[ 2 ] 1 25 1 25 2 5 2 5 Program Counter 3 9 Program Counter 3 9 44 … … 48 … … FFFFFFFF 0 FFFFFFFF 0 Instructions Instructions Register Data Register Data Instructions: 0 0 Instructions: 0 0 “The vocabulary of commands” 1 3 “The vocabulary of commands” 1 3 Specify how to operate on state 2 1 Specify how to operate on state 2 5 3 3 3 3 Example: … … Example: … … 40: add r1 = r2 + r3 31 0 40: add r1 = r2 + r3 31 0 44: sub r3 = r1 - r0 Address Data 44: sub r3 = r1 - r0 Address Data 48: store M[ r3 ] = r1 48: store M[ r3 ] = r1 0 0 0 0 52: load r25 = M[ 2 ] 52: load r2 = M[ 2 ] 1 25 1 25 2 5 2 5 Program Counter 3 3 Program Counter 3 3 52 … … 52 … … FFFFFFFF 0 FFFFFFFF 0 Instructions Assembly Instructions and C Register Data main() { Note: 0 0 int a = 15, b = 1, c = 2; 1. Insts Executed in Order 1 3 2. Addressing Modes 2 5 add r1 = r2 + r3 a = b + c; /* a gets 3 */ 3 3 … … Example: sub r3 = r1 - r0 c = a; /* c gets 3 */ 31 0 40: add r1 = r2 + r3 44: sub r3 = r1 - r0 Address Data store M[ r3 ] = r1 *(int *)c = a; 48: store M[ r3 ] = r1 /* M[c] = a */ 0 0 52: load r2 = M[ 2 ] 1 25 load r2 = M[ 2 ] b = *(int *)(2); 2 5 /* b gets M[2] */ Program Counter 3 3 } 52 … … FFFFFFFF 0 Branching Suppose we could only execute instructions in sequence. Recall from our example: 40: add r1 = r2 + r3 44: sub r3 = r1 - r0 48: store M[ r3 ] = r1 52: load r2 = M[ 2 ] • In a decent desktop machine, how long would the longest program stored in main memory take? • Assume: • 1 instruction per cycle • An instruction is encoded in 4 bytes (32 bits) 33 34 Therefore… Unconditional Branches • Some instructions must execute more than once • Unconditional branches always update the PC • PC must be updated • AKA: Jump instructions Example: Example: 40: add r1 = r2 + r3 40: add r1 = r2 + r3 44: sub r3 = r1 - r0 44: sub r3 = r1 - r0 48: store M[ r3 ] = r1 48: store M[ r3 ] = r1 52: load r2 = M[ 2 ] 52: load r2 = M[ 2 ] 56: PC = 40 56: jump 40 • How long with the program take? 35 36 Conditional Branch Conditional Branch • Conditional Branch sometimes updates PC • What does this look like in C? • AKA: Branch, Conditional Jump • Example “ ” • Example 10: Hello\n ; data in memory 36: arg1 = 10 ; argument memory address is 10 40: r1 = 10 40: r1 = 10 44: r1 = r1 - 1 44: r1 = r1 - 1 48: branch r1 > 0, 44 if r1 is greater than 0, PC = 44 48: call printf ; printf(arg1) 52: halt 52: branch r1 > 0, 44 56: halt • How long will this program take? Details about red instructions/data next time… 37 38 Indirect Branches Branch Summary • Reduce, Reuse, Recycle (instructions) • Branch address may also come from a register • Branch instructions update state • AKA: Indirect Jump Registers 0 1 2 3 4 5 6 7 Example: 0000 0788 B700 0010 0401 0002 0003 00A0 PC 8 9 A B C D E F 40: add r1 = r2 + r3 10 0000 0788 B700 0010 0401 0002 0003 00A0 44: sub r3 = r1 - r0 Main Memory 48: store M[ r3 ] = r1 ? 00: 0000 0000 0000 0000 0000 0000 0000 0000 08: 0000 0000 0000 0000 0000 0000 0000 0000 52: load r2 = M[ 2 ] 10: 9222 9120 1121 A120 1121 A121 7211 0000 56: jump r4 18: 0000 0001 0002 0003 0004 0005 0006 0007 20: 0008 0009 000A 000B 000C 000D 000E 000F 60: halt 28: 0000 0000 0000 FE10 FACE CAFE ACED CEDE .