4.1 CS356 Unit 4 Intro to x86 Instruction Set 4.2 Why Learn Assembly • To understand something of the limitation of the HW we are running on • Helpful to understand performance • To utilize certain HW options that high-level languages don't allow (e.g. operating systems, utilizing special HW features, etc.) • To understand possible security vulnerabilities or exploits • Can help debugging 4.3 Compilation Process CS:APP 3.2.2 void abs(int x, int* res) • Demo of assembler { if(x < 0) *res = -x; – $ g++ -Og -c -S file1.cpp else *res = x; • Demo of hexdump } Original Code – $ g++ -Og -c file1.cpp – $ hexdump -C file1.o | more Disassembly of section .text: 0000000000000000 <_Z3absiPi>: 0: 85 ff test %edi,%edi 2: 79 05 jns 9 <_Z3absiPi+0x9> • Demo of 4: f7 df neg %edi 6: 89 3e mov %edi,(%rsi) 8: c3 retq objdump/disassembler 9: 89 3e mov %edi,(%rsi) b: c3 retq – $ g++ -Og -c file1.cpp Compiler Output – $ objdump -d file1.o (Machine code & Assembly) Notice how each instruction is turned into binary (shown in hex) 4.4 Where Does It Live • Match (1-Processor / 2-Memory / 3-Disk Drive) where each item resides: – Source Code (.c/.java) = 3 – Running Program Code = 2 – Global Variables = 2 – Compiled Executable (Before It Executes) = 3 – Current Instruction Being Executed = 1 – Local Variables = 2 (1) Processor (2) Memory (3) Disk Drive 4.5 BASIC COMPUTER ORGANIZATION 4.6 Processor • Performs the same 3-step process over and over again – Fetch an instruction from Processor Arithmetic 3 Add the memory Circuitry specified values Decode 2 It’s an ADD – Decode the instruction Circuitry • Is it an ADD, SUB, etc.? 1 Fetch – Execute the instruction Instruction System Bus • Perform the specified operation • This process is known as the ADD SUB Instruction Cycle CMP Memory 4.7 Processor CS:APP 1.4 • 3 Primary Components inside a processor – ALU – Registers – Control Circuitry • Connects to memory and I/O via address, data, and control buses (bus = group of wires) Bus Processor Memory PC/IP 0 Addr Control 0 op. 1 ALU 2 out ADD, in1 SUB, Data 3 AND, 4 OR in2 R0-R31 5 Control 6 4.8 Arithmetic and Logic Unit (ALU) • Digital circuit that performs arithmetic operations like addition and subtraction along with logical operations (AND, OR, etc.) Processor Memory Addr ADD 0 op. 1 ALU 0x0123 2 out ADD, in1 SUB, Data 3 0x0579 0x0456 AND, 4 OR in2 5 Control 6 4.9 Registers • Recall memory is SLOW compared to a processor • Registers provide fast, temporary storage locations within the processor Processor Memory PC/IP Addr 0 op. 1 ALU 0x0123 2 out ADD, in1 SUB, 0x0456 Data 3 AND, 4 OR in2 R0-Rn-1 5 Control 6 4.10 General Purpose Registers • Registers available to software instructions for use by the programmer/compiler • Programmer/compiler is in charge of using these registers as inputs (source locations) and outputs (destination locations) Processor Memory PC/IP Addr 0 op. 1 ALU 2 out ADD, in1 SUB, Data 3 AND, 4 OR in2 R0-Rn-1 5 Control 6 4.11 What if we didn’t have registers? • Example w/o registers: F = (X+Y) – (X*Y) – Requires an ADD instruction, MULtiply instruction, and SUBtract Instruction – w/o registers • ADD: Load X and Y from memory, store result to memory • MUL: Load X and Y again from mem., store result to memory • SUB: Load results from ADD and MUL and store result to memory • 9 memory accesses Processor Memory PC/IP Addr 0 X op. 1 Y ALU 2 out ADD, in1 SUB, Data 3 F AND, 4 OR in2 R0-Rn-1 5 Control 6 4.12 What if we have registers? • Example w/ registers: F = (X+Y) – (X*Y) – Load X and Y into registers – ADD: R0 + R1 and store result in R2 – MUL: R0 * R1 and store result in R3 – SUB: R2 – R3 and store result in R4 – Store R4 back to memory – 3 total memory access Processor Memory PC/IP Addr 0 X op. 1 Y ALU 2 out ADD, in1 X SUB, Y Data 3 F AND, 4 OR in2 R0-Rn-1 5 Control 6 4.13 Other Registers • Some bookkeeping information is needed to make the processor operate correctly • Example: Program Counter/Instruction Pointer (PC/IP) Reg. – Recall that the processor must fetch instructions from memory before decoding and executing them – PC/IP register holds the address of the next instruction to fetch Processor Memory PC/IP Addr 0 op. 1 ALU 2 out ADD, in1 SUB, Data 3 AND, 4 OR in2 R0-Rn-1 5 Control 6 4.14 Fetching an Instruction • To fetch an instruction – PC/IP contains the address of the instruction – The value in the PC/IP is placed on the address bus and the memory is told to read – The PC/IP is incremented, and the process is repeated for the next instruction Processor PC/IP = Addr = 0 Memory PC/IP 0 Addr 0 inst. 1 op. 1 inst. 2 ALU Data = inst.1 machine code 2 inst. 3 out ADD, in1 SUB, Data 3 inst. 4 AND, 4 inst. 5 OR in2 Control = Read R0-Rn-1 … Control FF 4.15 Fetching an Instruction • To fetch an instruction – PC/IP contains the address of the instruction – The value in the PC/IP is placed on the address bus and the memory is told to read – The PC/IP is incremented, and the process is repeated for the next instruction Processor PC/IP = Addr = 1 Memory PC/IP 1 Addr 0 inst. 1 op. 1 inst. 2 ALU Data = inst.2 machine code 2 inst. 3 out ADD, in1 SUB, Data 3 inst. 4 AND, 4 inst. 5 OR in2 Control = Read R0-Rn-1 … Control FF 4.16 Control Circuitry • Control circuitry is used to decode the instruction and then generate the necessary signals to complete its execution • Controls the ALU • Selects Registers to be used as source and destination locations Processor Memory PC/IP 0 Addr Control 0 inst. 1 op. 1 inst. 2 ALU 2 inst. 3 out ADD, in1 SUB, Data 3 inst. 4 AND, 4 inst. 5 OR in2 R0-Rn-1 … Control FF 4.17 Control Circuitry • Assume 0x0201 is machine code for an ADD instruction of R2 = R0 + R1 • Control Logic will… – select the registers (R0 and R1) – tell the ALU to add – select the destination register (R2) Processor 0 Memory PC/IP 0 Addr Control 0 0201 ADD 1 inst. 2 0x0123 0201 2 inst. 3 out ALU in1 0x0456 Data 3 inst. 4 ADD 0x0579 4 in2 inst. 5 R0-Rn-1 … Control FF 4.18 Summary • Registers are used for fast, temporary storage in the processor – Data (usually) must be moved into registers • The PC or IP register stores the address of the next instruction to be executed – Maintains the current execution location in the program 4.19 UNDERSTANDING MEMORY 4.20 Memory and Addresses • Set of cells that each store a group of bits Address Data A[0] – Usually, 1 byte (8 bits) per 0 11010010 cell 1 01001011 … 2 10010000 • Unique address (number) A[n-1] assigned to each cell Inputs Address 3 11110100 – Used to reference the value 4 01101000 5 11010001 in that location D[0] • Data and instructions are … … both stored in memory and D[7] are always represented as a FFFF 00001011 string of 1’s and 0’s Inputs/Outputs Data Memory Device 4.21 Reads & Writes 0 11010010 • Memories perform 2 operations 2 1 01001011 Addr. – Read: retrieves data value in a 10010000 2 10010000 Data particular location (specified using 3 11110100 the address) Processor 4 01101000 – Write: changes data in a location 5 11010001 Read … to a new value Control • To perform these operations a set of address, data, and control FFFF 00001011 wires are used to talk to the A Read Operation memory 0 11010010 1 01001011 5 – Note: A group of wires/signals is Addr. 2 10010000 referred to as a ‘bus’ 00000110 Data 3 11110100 – Thus, we say that memories have 4 01101000 Processor an address, data, and control bus. 5 00000110 … Write Control FFFF 00001011 System Bus (address, data, control wires) A Write Operation 4.22 Memory vs. I/O Access • Processor performs reads and writes to communicate with memory and I/O devices – I/O devices have memory locations that contain data that the processor can access – All memory locations (be it RAM or I/O) have unique addresses which are used to identify them – The assignment of memory addresses is known Processor Memory as the physical memory map 0 code … 0x3ffffff data A D C Video Interface 800 FE may 8000000 FE FE signify a … WRITE white dot at a particular 01 location ‘a’ = 61 hex Keyboard in ASCII Interface 4000000 61 4.23 Address Space Size and View Logical View 0xf_ffff_ffff Memory Processor I/O Dev 2 I/O Dev 1 System (Addr. + Data) Bus OS Code (Addr = 36-39 bits, Data = 64) OS Stack Logical Address & Data bus widths = 64-bits User Stack Globals • Most computers are byte-addressable Code movl %rax,(%rdx) – Each unique address corresponds to 1-byte of addl %rcx,%rax ... memory (so we can access char variables) 0x0 • Address width determines max amount of Logical view of memory address/memory space – Every byte of data has a unique address – 32-bit addresses => 4 GB address space – 36-bit address bus => 64 GB address space 4.24 Data Bus & Data Sizes • Moore's Law meant we could build systems with more transistors Processor Data Bus • More transistors meant greater bit-widths Width – Just like more physical space allows for wider Intel 8088 8-bit roads/freeways, more transistors allowed us to move to 16-, 32- and 64-bit circuitry inside the processor Intel 8086 16-bit • To support smaller variable sizes (char = 1-byte) we Intel 80386 32-bit still need to access only 1-byte of memory per Intel Pentium 64-bit access, but to support int and long ints we want to access 4- or 8-byte chunks of memory per access Processor • Thus the data bus (highway connecting the processor and memory) has been getting wider (i.e.