2021.11.10 compilation examination

Just make a note of it , It's more fragmented

The difference between memory addressing methods

2. Instructions MOV AX, [BX+DI] The addressing mode of the source operand in is :( d )（5 branch ）
A. Register indirection
B. Register addressing
C. Register Relative Addressing
D. Base addressing

① For memory addressing
The registers that can be addressed are
Base register ：bp and bx
Index register : bi and si
Two index registers cannot be used at the same time
② Register indirection means that only the contents of registers are used as offset addresses ：[bx]
③ In register addressing “ relative ” Addressing means adding an offset to the register ：[bx+666]

Two symbolic extension instructions

3. Execution instruction “MOV AL, 85H” and “CBW” after ,AX The contents of the register are :( c )（5 branch ）
A. 85H
B. 0085H
C. FF85H
D. 85F0H

① There are two symbolic extension instructions :cbw and cwd
According to my own understanding
cbw:convert Byte to Word
This command can put al This number extends to ax
cwd:convert Word to Double
This command can put ax This number extends to dx.ax
therefore , The essence of symbolic extension instruction is actually type conversion
② The symbol bit expansion method is relatively simple , Look directly at the binary encoding mode in this register , If the highest order is 1, Just expand 1,0 Just 0

db Instruction definition string

4. Define a variable in the data segment VAR DB ‘12345678H’, Execution instruction “MOV CL,VAR+3” after ,CL The contents of the register are :(b)（5 branch ）
A. 12H
B. 34H
C. 56H
D. 78H

①db If you want to define numbers later , The size cannot be greater than one byte . If you define a string , Then the string will be in the order from left to right , Store characters from low address to high address , One byte per character , that ,var+3 after , Just pointing ’4’ This character , This is the time , Check it out ascii surface , Symbol ‘4’ Corresponding ascii Code is 34h

Take out the segment address of a segment

6. What can be used as a programmer's custom identifier in assembly language is :( a )（5 branch ）
A. done
B. idiv
C. xlat
D. @data

@data Get the segment address of the data segment . such as ,mov ax,@data
A more general approach is
adopt mov bx,offset var1 Get the offset address
adopt mov ds,seg var1 Get segment address

SF and OF Sign a

7. if AX=1234H, Execution instruction “ADD AX,7802H” after ,AX Register contents and flag bits CF,SF,ZF,OF The state of is :(b)（5 branch ）
A. AX=8A36H CF=0 SF=1 ZF=1 OF=0
B. AX=8A36H CF=0 SF=1 ZF=0 OF=1
C. AX=9036H CF=0 SF=1 ZF=1 OF=0
D. AX=9036H CF=0 SF=1 ZF=0 OF=1

①OF: overflow flag. When considering this flag bit , You need to treat the addend and the addend as signed numbers , in other words , Both the addend and the addend exist in the form of complement , After adding , The result is beyond 16 The largest positive number of bit complement ：0fff, It means overflow ,OF Set to 1. in other words , When considering the value of a flag bit , It needs to be considered from the perspective of the computer rather than the practical significance of the data you store .
②SF：sign flag sign indicator . What is the highest result ,SF What is it?
③SF and OF It is meaningful when performing symbolic operations , But there will also be value transformation when performing unsigned operations .
④PF： Parity flag bit ,PF This flag bit will be combined with 1 The number of is odd , That is, if there are odd numbers in the operation result 1, be pf by 0

Subtraction instruction with borrow sbb

12. if AX=1020H, BX=1200H, CF=1, Then the command “SBB AX, BX” The execution result of is ：（b）（5 branch ）
A. 0FE20H
B. 0FE1FH
C. 0FE21H
D. 01DFH

sbb The instruction needs to take into account the borrowing left in the last calculation , meanwhile , This calculation also needs to be modified if it is borrowed CF Value . Theoretically , The result is ax-bx-cf, however ax Not enough to reduce , So we have to borrow again , namely (F+1)020-1200-1=FE1F. Because this calculation borrowed , After that , set up cf=1

The first of the registers x Which bit refers to

14. It is known that （AL）=56H, If the AL Of the 0, 2, 6 Bit changes, but other bits remain unchanged , Then use the command （ ） AL, BL; among （BL）=( a ).（5 branch ）
A. XOR 45H
B. OR 45H
C. XOR 0A2H
D. AND 0BBH

From right to left, the first is 0 position

The difference between immediate addressing and direct addressing

① Immediate addressing means that a number is stored directly in the instruction
② Direct addressing is a kind of memory addressing , Directly locate the memory unit by a symbol or offset , such as ,mov ax,var perhaps mov ax,[var]

Segment registers and mov Cooperation of instructions

① In the textbook , The abbreviation of segment register is seg
② In assembly language , The segment register has s Various registers at the end ds cs ss es
③ stay mov In the instruction , Segment registers cannot be combined with immediate addressing , That is, not allowed :mov seg,imm

dosbox What do you think of the flag bit in

Post it here in the experiment report

The influence of some common instructions on flag bits

① Memory of circular instructions ,rcl There is one of them. c, On behalf of this instruction, we need to consider CF, That is, this is a left loop instruction with carry
②rcl perhaps rol These loop instructions all affect CF, The numbers squeezed out of the cycle will be stored in CF in
③ Bit operation instruction and,or,xor These instructions will directly of and cf Set up 0, Cause unknown , The main reason is that different instructions have specific purposes when they are designed , The influence ability on the flag bit is different , So remember first , Prepare for the exam
④mov The instruction does not affect the flag bit , Mainly used to transmit data
⑤ utilize test The instruction tests whether a number is negative , Or judge whether it is zero . for example ,test ax,ax. because test The essence of instruction is and Instructions , The difference lies in test Instruction operands , But the impact zf and sf Sign a ,test ax,ax The result is ax In itself , If ax It's zero , Then it will be set zf=1, If ax It's a negative number , Then it will be set sf=1

Differences in addressing of various instructions

① Intuitively distinguish from data addressing , Namely , The instructions involved in data addressing are mov, The instructions involved in instruction addressing are various jump statements jmp loop wait
② The purpose of instruction addressing is to jump to a specific location , In essence, they are all modifications cs:ip The data of （cs:ip The content of is the address of the next instruction ）, As modified cs:ip The way , It can be divided into near transfer , Short transfer , Far away
③ According to whether to modify cs The content of , It is also divided into intra segment addressing and inter segment addressing
④ Near transfer is called relative addressing , It is also an intra segment addressing , for example jmp labl1(labl1 It's a label , such as labl1:mov ax,1) Or is it jmp near ptr lablq, Or is it loop Instructions , The practice of near transfer is in the current ip Add a displacement to the value , The limitation is that the displacement is in +127~-128 byte
⑤ Short transfers may be direct addressing or register indirection or memory indirection , It is also an intra segment addressing , for example jmp short ptr labl1 perhaps jmp ax perhaps jmp [ax], The method of short transfer is to directly label labl1 The effective address of or the contents of the register are assigned to ip
⑥ The method of far transfer is to modify at the same time cs:ip, It is an inter segment addressing , for example jmp far ptr labl1
⑦ The basis of various classifications is different , In fact, when you remember, you don't have to re classify in a classification , Just remember the basis , The above is just to deepen our understanding

Stack pass parameter considerations

① You want to add some entry parameters before calling a subroutine , It's calling call Use... Before the command push Instructions , It should be noted that ,call The instruction itself pushes something into the stack , The return address
② Generally speaking , We use in subroutines ret The command returns to the main program , What this instruction does for us is , Pop up return address , Use this return address to set cs:ip, But don't forget at this time , The top of the stack is still the entry parameter put in before , So generally speaking , When using the stack to pass parameters , Need to use ret i16 Command to return , This instruction is equivalent to ,ret after ,sp+i16, That is, lower the pointer at the top of the stack to the position where no entry parameter is added