Assembly language programming final examination site

Little knowledge

1. 10 times of 2 is 1k

2. When the address bus is 20 bits, the maximum memory is 1m. Reason: the 20th power of 2 1024k=1M

3. A variable is the first address that the storage unit points to

Segment register offset address

• The CS code segment IP # CS can only be the source operand, not the destination operand MOV CS, AX; error

• DS data segment BX, SI, DI or a 16 digit

• SS stack segment SP or BP

• ES additional segment

Segment base value (16 bits) shifts 4 bits to the left (i.e. 4-bit binary 0 is added at the end) = segment base address

EA (effective address) = base address + (index value X scale factor) + displacement

Physical address = EA + segment base address

other

1. All the instructions that can be executed by a computer are called the instruction system or instruction set of the computer. The specific contents include instruction format, addressing mode and instruction type

The instruction system layer is the interface between hardware and software. The task of hardware is to execute instructions, and the program is embodied as instruction sequence

1. 8086CPU is a 16 bit microprocessor. The internal and external data bus of CPU is 16 bits. One word (16 bits) data can be transmitted in a bus cycle, and the address bus is 20 bits, which can address 1MB main memory space
   The serial sequence cannot read the next instruction from main memory until one instruction is executed

The main memory capacity of 8086 / 8088 CPU is 1MB, and its main memory unit address is 20 bits, while the internal registers and data paths of CPU are 16 bits. How to expand the 16 bit address code to 20 bits?

Solution: divide 1MB main memory space into several segments, and the maximum length of each segment is 64KB. Accordingly, a register is set in the BIU to store the upper 16 bits of the 20 bit starting address, which is called the segment base value. The EU or instruction pointer IP provides the offset address (also known as offset) in the segment: the byte distance between a main memory unit and the segment base address of the segment

Address bus 20 bits 00000H-FFFFFH, byte storage, word storage, double word storage

Effective address: it is usually calculated by the CPU according to the addressing mode of the instruction or provided by the instruction pointer IP

Logical operation instruction

DEST； Destination operand SRC: source operand

• AND logic AND instruction
  • Instruction format: AND DEST, SRC
  • 0 is 0, or 1 is 1
• OR logic OR instruction
  • Instruction format: OR DEST, SRC
  • If one is 1, the result is 1
• XOR logical XOR instruction
  • Instruction format: XOR DEST, SRC
  • 0 for the same and 1 for the different
• NOT logical non instruction
  • Instruction format: NOT DEST
  • Reverse

Note: in the above operations, CF and OF will be set to 0 except NOT

Arithmetic instructions

1. ADD addition

   • Instruction format: ADD DEST, SRC
   • The two numbers are added and the result is stored in DEST

2. MOV transfer instruction

   • Instruction format: MOV DEST, SRC
   • Copy the operands in SRC to DEST

3. XCHG exchange instruction

   • Instruction format: MOV DEST, SRC
   • Exchange the contents of the source and destination operands with each other

4. Subtraction instruction

   1. SUB instruction

      • Instruction format: SUB DEST, SRC
      • Completes reducing the de source operand from the destination operand, that is: SEST = (SEST) - (SRC)

   2. DEC instruction

      • Instruction format: DEC DEST
      • Function: self minus one, DEST = (DEST) - 1

   3. CMP instruction

      • Instruction format: CMP DEST, SRC
      • The difference between the two operands does not retain the result, but only the changed flag bits: OF,SF,ZF,AF,PF, etc

Displacement command

• SHL logic shift left
• SHR logic shift right
• Shift SAL arithmetic left
• SAR arithmetic shift right
• ROL cycle shift left
• ROR cycle shift right
• RCL with carry cycle shift left
• RCR with carry cycle shift right

Instruction format:

RCL DEST, COUNT

Assembler case

Data segment definition format:

DATA SEGMENT ；Data segment DATA ENDS

Stack segment definition format:

STACK1 SEGMENT PARA STACK ;stack segment DW 20H DUP (0) STACK1 ENDS

CODE SEGMENT ;Code snippet ASSUME CS: CODE , DS:DATA,ES:EXTERA;Assume the storage location of each segment START: MOV AX, DATA ;Segment based refers to DS MOV DS,AX Instruction sequence MOV AH,4CH ;Call 4 CH Function, 4 CH Function number INT 21H ;Termination procedure CODE ENDS END START

Assembly language statement format (case):

calculation:
F = ( 10 ∗ ( X + Y ) − 3 ∗ ( Z − 1 ) ) / 2 F = (10 * (X + Y)-3 * (Z-1))/2 F=(10∗(X+Y)−3∗(Z−1))/2
The results are stored in FUN

DATA SEGMENT;Set data segment VARX DW 123H;Define variables x VARY DW 456H;Y VARZ DW 789H;Z FUN DW ? Set stack DATA ENDS STACK1 SEGMENT PARA STACK； DW 20H DUP(0) STACK1 ENDS CODE SEGMENT；Set code snippet ASSUME CS: CODE, DS: DATA, SS: STACK1 ;Assume the storage location of each segment STAET: MOV AX, DATA ;Segment based refers to DS MOV DS,AX MOV AX,VARX ;Take the variable and copy it to AX in ADD AX,VARY ;AX = (X+Y) MOV BX,AX ;BX = AX = (X+Y) SAL AX,1 ;AX = 2 * (X+Y) Move the arithmetic left, equivalent to X2 SAL AX,1 ;AX = 4 * (X+Y) ADD AX,BX ;AX = 5 * (X+Y) SAL AX,1 ;AX = 10 * (X+Y) MOV BX,VARZ ;Take variable Z DEC BX ;BX = BX-1 = Z-1 MOV CX,BX ;CX = BX = Z-1 SAL BX,1 ;CX = 2 * (Z-1) ADD BX,CX ;BX = 3 * (Z-1) SUB AX,BX ;AX = 10 * (X+Y) - 2 * (Z-1) SAR AX,1 ;ax = /2 Move the arithmetic right, equivalent to /2 MOV FUN,AX ;Storage results MOV AH,4CH ;Call 4 CH Function, 4 CH Function number INT 21H ;Termination procedure ODE ENDS END START

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