02计算机组成—汇编语言编程

首先介绍了高级语言与汇编语言编程的区别、对汇编语言的理解、可视化指令执行、汇编语言说明、机器语言编程,然后介绍了怎样区别指令和数据,介绍了比特、字节、字的概念,寄存器和内存,最后演示了简单的汇编语言编程。
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1.Assembly Language Programming 1

2.High-level vs. Assembly language Consider the following statements 1. a=x+y–z 2. if x > y then x:= x + y else x:= x - y HLL (High Level Language) programs are machine independent. They are easy to learn, easy to use, and convenient for managing complex tasks. Assembly language programs are machine specific. It is the language that the processor “directly” understands. HLL Compiler Assembly Language 2

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4.Understanding Assembly Language Let us begin with data representation. How to represent • Signed integers • Fractions • Alphanumeric characters Review • Floating point numbers • Pictures? Memory 01001011 11011010 1 00 1 1 0 0 0 Can you read 0o00 the contents of these memory cells? 4

5.Visualizing instruction execution (The main concept is register-transfer operation. registers Memory 0 500 x r0 1 24 y r1 ALU -32 2 z r2 0 3 a r3 Address data processor A register is a fast storage within the CPU load x into r1 load y into r2 a = x + y - z load z into r0 r3 ← r1 + r2 r0 ← r3 – r0 store r0 into a 5

6.Assembly language instructions for a hypothetical machine (not MIPS) Load x, r1 Load y, r2 Load z, r0 Add r3, r1, r2 Sub r0, r3, r0 Store r0, a Each processor has a different set of registers, and different assembly language instructions. The assembly language instructions of Intel Pentium and MIPS are completely different. Motorola 68000 has 16 registers r0-r15 MIPS has 32 registers r0-r31 Pentium has 8 general purpose & 6 segment registers. 6

7.Binary or Machine Language program Both program and data are represented using only 0’s and 1’s inside a computer. Here is a sample: 0 31 010100 11001100…000 0 Load address of x Operation code 0 31 110010 11 01 10 Add r3 r1 r2 unused These are instruction formats. Each instruction has a specific format. 7

8.Can we distinguish program from data? Program Both are bit strings. Data Indistinguishable. MEMORY Normally, the programmer has to tell the machine (or use some convention) to specify the address of the first instruction. Incorrect specification will lead to errors, and the program is most likely to crash. 8

9.Bits, bytes, words Bit: 0, 1 Byte: string of 8 bits. Each byte has an address. Word: one or more bytes (usually 2 or 4 or 8). 0 01010000 1 11110000 word 0 0000000 2 11111111 3 00001111 4 10111011 5 word 1 00111100 6 00000111 7 9

10. 0 Word 0 Word 1 4 Word 2 8 12 Word 3 Byte order in a word Big Endian order [byte 0, byte 1, byte 2, byte 3] Little Endian order [byte 3, byte2, byte 1, byte 0] 10

11.Registers vs. memory Data can be stored in registers or memory locations. Memory access is slower (takes approximately 50 ns) than register access (takes approximately 1 ns or less). To increase the speed of computation it pays to keep the variables in registers as long as possible. However, due to technology limitations, the number of registers is quite limited (typically 8-64). MIPS registers MIPS has 32 registers r0-r31. The conventional use of these registers is as follows: 11

12.register assembly name Comment r0 $zero Always 0 r1 $at Reserved for assembler r2-r3 $v0-$v1 Stores results r4-r7 $a0-$a3 Stores arguments r8-r15 $t0-$t7 Temporaries, not saved r16-r23 $s0-$s7 Contents saved for later use r24-r25 $t8-$t9 More temporaries, not saved r26-r27 $k0-$k1 Reserved by operating system r28 $gp Global pointer r29 $sp Stack pointer r30 $fp Frame pointer r31 $ra Return address 12

13.Example assembly language programs Example 1 f=g+h–i Assume that f, g, h, i are assigned to $s0, $s1, $s2, $s3 add $t0, $s1, $s2 # register $t0 contains g + h sub $s0, $t0, $s3 #f=g+h–i Example 2. g = h + A[8] Assume that g, h are in $s1, $s2. A is an array of words, the elements are stored in consecutive locations of the memory. The base address is stored in $s3. lw t0, 32($s3) # t0 gets A[8], 32= 4x 8 add $s1, $s2, $t0 # g = h + A[8] 13

14.Machine language representations Instruction “add” belongs to the R-type format. opcode rs rt rd shift amt function 6 5 5 5 5 6 src src dst add $s1, $s2, $t0 will be coded as 0 18 8 17 0 32 6 5 5 5 5 6 The function field is an extension of the opcode, and they together determine the operation. Note that “sub” has a similar format. 14