Assembly Language Professor Jennifer Rexford COS 217 1

Goals of Today’s Lecture • Assembly language o. Between high-level language and machine code o. Why learn assembly-language programming? • Computer architecture o. Central Processing Unit (CPU) o. Fetch-decode-execute cycle • C code vs. assembly code o. Small IA 32 example to illustrate the relationship to C o. Illustrates storing, manipulating, comparing, and jumping 2

Three Levels of Languages 3

High-Level Language • Make programming easier by describing operations in a natural language • Increase the portability of the code • One line may involve many low-level operations • Examples: C, C++, Java, Pascal, … count = 0; while (n > 1) { count++; if (n & 1) n = n*3 + 1; else n = n/2; } 4

Assembly Language • Tied to the specifics of the underlying machine movl. loop: cmpl jle addl movl andl je movl addl jmp • Commands and names to make the code readable and writeable by humans • Hand-coded assembly code may be more efficient • E. g. , IA 32 from Intel . else: sarl. endif: jmp. endloop: $0, %ecx $1, %edx. endloop $1, %ecx %edx, %eax $1, %eax. else %edx, %eax, %edx $1, %edx. endif $1, %edx. loop 5

Machine Language • Also tied to the underlying machine 0000 0000 • What the computer sees and deals with • Every command is a sequence of one or more numbers • All stored in memory on the computer, and read and executed 0000 0000 9222 9120 1121 A 121 7211 0000 0001 0002 0003 0004 0005 0006 0007 0008 0009 000 A 000 B 000 C 000 D 000 E 000 F 0000 FE 10 FACE CAFE ACED CEDE 1234 5678 9 ABC DEF 0 0000 F 00 D 0000 EEEE 1111 0000 B 1 B 2 F 1 F 5 0000 0000 • Unreadable by humans 6

Why Learn Assembly Language? • Write faster code (even in high-level language) o. By understanding which high-level constructs are better o… in terms of how efficient they are at the machine level • Understand how things work underneath o. Learn the basic organization of the underlying machine o. Learn how the computer actually runs a program o. Design better computers in the future • Some software is still written in assembly language o. Code that really needs to run quickly o. Code for embedded systems, network processors, etc. 7

Why Learn Intel IA 32 Assembly? • Program natively on our computing platform o. Rather than using an emulator to mimic another machine • Learn instruction set for the most popular platform o. Most likely to work with Intel platforms in the future • But, this comes at some cost in complexity o. IA 32 has a large and varied set of instructions o. Arguably more than are really useful in practice • Fortunately, you won’t need to use everything 8

Computer Architecture 9

A Typical Computer CPU Memory . . . CPU Chipset I/O bus ROM Network 10

Von Neumann Architecture • Central Processing Unit o. Control unit – Fetch, decode, and execute o. Arithmetic and logic unit – Execution of low-level operations o. General-purpose registers – High-speed temporary storage o. Data bus – Provide access to memory CPU Control Unit • Memory o. Store instructions o. Store data ALU Registers Data bus Random Access Memory (RAM) 11

Control Unit: Instruction Pointer • Stores the location of the next instruction o. Address to use when reading from memory • Changing the instruction pointer (EIP) o. Increment by one to go to the next instruction o. Or, load a new value to “jump” to a new location EIP 12

Control Unit: Instruction Decoder • Determines what operations need to take place o. Translate the machine-language instruction • Control what operations are done on what data o. E. g. , control what data are fed to the ALU o. E. g. , enable the ALU to do multiplication or addition o. E. g. , read from a particular address in memory ALU 13

Registers • Small amount of storage on the CPU o. Can be accessed more quickly than main memory • Instructions move in and out of registers o. Loading registers from main memory o. Storing registers to main memory • Instructions manipulate the register contents o. Registers essentially act as temporary variables o. For efficient manipulation of the data • Registers are the top of the memory hierarchy o. Ahead of cache, main memory, disk, tape, … 14

Keeping it Simple: All 32 -bit Words • Simplifying assumption: all data in four-byte units o. Memory is 32 bits wide o. Registers are 32 bits wide EAX EBX • In practice, can manipulate different sizes of data 15

C Code vs. Assembly Code 16

Example: Kinds of Instructions • Storing values (in registers) count = 0; while (n > 1) { count++; if (n & 1) n = n*3 + 1; else n = n/2; } o count = 0 on • Arithmetic and logic operations o Increment: count++ o Multiply: n * 3 o Divide: n/2 o Logical AND: n & 1 • Checking results of comparisons o while (n > 1) o if (n & 1) • Jumping o To the end of the while loop (if “n > 1”) o Back to the beginning of the loop o To the else clause (if “n & 1” is 0) 17

Example: Variables in Registers count = 0; while (n > 1) { count++; if (n & 1) n = n*3 + 1; else n = n/2; } Registers n %edx count %ecx Referring to a register: percent sign (“%”) 18

Example: Immediate and Register Addressing count=0; while (n>1) { count++; if (n&1) n = n*3+1; else n = n/2; } movl $0, %ecx addl $1, %ecx Read directly from the instruction Referring to a constant: dollar sign (“$”) written to a register 19

Example: Immediate and Register Addressing count=0; while (n>1) { count++; if (n&1) n = n*3+1; else n = n/2; } movl andl %edx, %eax $1, %eax Computing intermediate value in register EAX 20

Example: Immediate and Register Addressing count=0; while (n>1) { count++; if (n&1) n = n*3+1; else n = n/2; } movl addl %edx, %eax, %edx $1, %edx Adding n twice is cheaper than multiplication! 21

Example: Immediate and Register Addressing count=0; while (n>1) { count++; if (n&1) n = n*3+1; else n = n/2; } sarl $1, %edx Shifting right by 1 bit is cheaper than division! 22

Example: Changing Program Flow • Cannot simply run next instruction count=0; while (n>1) { count++; if (n&1) n = n*3+1; else n = n/2; } o. Check result of a previous operation o. Jump to appropriate next instruction • Flags register (EFLAGS) o. Stores the status of operations, such as comparisons o. E. g. , last result was positive, negative, zero, etc. • Jump instructions o. Load new address in instruction pointer • Example jump instructions o. Jump unconditionally (e. g. , “}”) o. Jump if zero (e. g. , “n&1”) o. Jump if greater/less (e. g. , “n>1”) 23

Example: Jump and Labels movl. loop: count=0; while (n>1) { count++; if (n&1) n = n*3+1; else n = n/2; } cmpl jle addl movl andl je movl addl jmp . else: sarl. endif: jmp. endloop: $0, %ecx $1, %edx. endloop $1, %ecx %edx, %eax $1, %eax. else %edx, %eax, %edx $1, %edx. endif $1, %edx. loop 24

Example: Jump and Labels movl. loop: count=0; while(n>1) { count++; if (n&1) n = n*3+1; else “if” n = n/2; } “else” block cmpl jle addl movl andl je movl addl block addl jmp. else: sarl. endif: jmp. endloop: $0, %ecx $1, %edx. endloop $1, %ecx %edx, %eax $1, %eax. else %edx, %eax, %edx $1, %edx. endif $1, %edx. loop 25

Example: Making More Efficient? ? movl. loop: count=0; while(n>1) { count++; if (n&1) n = n*3+1; else n = n/2; } Replace with “jmp. loop” cmpl jle addl movl andl je movl addl jmp. else: sarl. endif: jmp. endloop: $0, %ecx $1, %edx. endloop $1, %ecx %edx, %eax $1, %eax. else %edx, %eax, %edx $1, %edx. endif $1, %edx. loop 26

n %edx count %ecx Complete Example movl. loop: count=0; while (n>1) { count++; if (n&1) n = n*3+1; else n = n/2; } cmpl jle addl movl andl je movl addl jmp . else: sarl. endif: jmp. endloop: $0, %ecx $1, %edx. endloop $1, %ecx %edx, %eax $1, %eax. else %edx, %eax, %edx $1, %edx. endif $1, %edx. loop 27

Reading IA 32 Assembly Language • Assembler directives: starting with a period (“. ”) o. E. g. , “. section. text” to start the text section of memory o. E. g. , “. loop” for the address of an instruction • Referring to a register: percent size (“%”) o. E. g. , “%ecx” or “%eip” • Referring to a constant: dollar sign (“$”) o. E. g. , “$1” for the number 1 • Storing result: typically in the second argument o. E. g. “addl $1, %ecx” increments register ECX o. E. g. , “movl %edx, %eax” moves EDX to EAX • Comment: pound sign (“#”) o. E. g. , “# Purpose: Convert lower to upper case” 28

Designing an Instruction Set 29

How Many Instructions to Have? • Need a certain minimum set of functionality o. Must to be able to represent any computation that can be expressed in a higher-level language • Benefits of having many instructions o. Direct implementation of many key operations o. Represent a line of C in one (or a few) lines of assembly • Disadvantages of having many instructions o. Larger opcode size o. More complex logic to implement complex instructions o. Hard to write compilers to exploit all of the instructions o. Hard to optimize the implementation of the CPU 30

CISC vs. RISC Complex Instruction Set Computer Reduced Instruction Set Computer (old fashioned, 1970 s style) (“modern”, 1980 s style) Examples: Vax (1978 -90) MIPS (1985 -? ) Motorola 68000 (1979 -90) Sparc (1986 -2006) 8086/80 x 86/Pentium (1974 -2025) IBM Power. PC (1990 -? ) ARM Instructions of various lengths, designed to economize on memory (size of instructions) IA 32 is a CISC architecture Instructions all the same size and all the same format, designed to economize on decoding complexity (and time, and power drain) 31

Next Time: More on Assembly Language • Machine code o. Encoding the operation and the operands o. Simpler MIPS instruction set as an example • Greater detail on IA 32 assembly language o. Different sizes of data (e. g. , byte, word, and extended) o. IA 32 instructions and addressing modes • Portions of memory o. Text, Data, BSS, Heap, and Stack • Calling functions o. Pushing and popping from the stack o. Arguments to a function and local variables o. Returning back to the calling function 32