MCS-012 SOLVED ASSIGNMENT FOR BCA/MCA Course Code: MCS-012 Course Title: Computer Organisation And Assembly Language Programming Assignment Number: MCS-012/Assignment/17-18 Last Date of Submission: 15 th Oct, 2017 (for Jul-2017 batch) : 15th April, 2018(for Jan-2018 batch)
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MCS-012 SOLVED ASSIGNMENT FOR BCA/MCA Ans. 4(a). Each string instruction may require a source operand, a destination operand or both. For 32-bit segments, string instructions use ESI and EDI registers to point to the source and destination operands, respectively. For 16-bit segments, however, the SI and the DI registers are used to point to the source and destination, respectively. Ther e are five basic instructions for processing strings. They are −
MOVS − This instruction moves 1 Byte, Word or Doubleword of data from memory location to another.
LODS − This instruction This instruction loads from memory. If the operand is of one byte, it is loaded into the AL register, if the operand is one word, it is loaded into the AX register and a doubleword is loaded into the EAX register.
STOS − This instruction stores data from regis ter (AL, AX, or EAX) to memory.
CMPS − This instruction compares two data items in memory. Data could be of a
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byte size, word or doubleword.
SCAS − This instruction compares the contents of a register (AL, AX or EAX) with the contents of an item in memory.
Each of the above instruction has a byte, word, and doubleword version, and string instructions can be repeated by using a repetition prefix. These instructions use the ES:DI and DS:SI pair of registers, where DI and SI registers contain valid offset addresses that refers to bytes stored in memory. SI is normally associated with DS (data segment) and DI is always associated with ES (extra segment).
DATA SEGMENTSTR DB ‘AXYBCSDEF$’ SUBSTR DB ‘BCS$’ LEN1 DB 0 LEN2 DB 0 MSG1 DB 10,13,’STRING IS : $’ MSG2 DB 10,13,’SUBSTRING IS : $’ MSG3 DB 10,13,’SUBSTRING IS FOUND AT POSITION POS ITION : $’
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MCS-012 SOLVED ASSIGNMENT FOR BCA/MCA POS DB -1 RTN DB ‘-1$’ DATA ENDS DISPLAY MACRO MSG MOV AH,9 LEA DX,MSG INT 21H ENDM CODE SEGMENT ASSUME CS:CODE,DS:DATA START: MOV AX,DATA MOV DS,AX DISPLAY MSG1 DISPLAY STR DISPLAY MSG2 DISPLAY SUBSTR LEA SI,STR NXT1: CMP [SI],’$’ JE DONE1 INC LEN1 INC SI JMP NXT1 DONE1: LEA DI,SUBSTR NXT2: CMP [DI],’$’ JE DONE2 INC LEN2 INC DI JMP NXT2 DONE2: DISPLAY MSG3 LEA SI,STR MOV AL,LEN1 SUB AL,LEN2 MOV CL,AL MOV CH,0 FIRST: INC POS
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MCS-012 SOLVED ASSIGNMENT FOR BCA/MCA MOV AL,[SI] CMP AL,SUBSTR[0] JE CMPR INC SI LOOP FIRST CMPR: INC SI MOV AL,[SI] CMP AL,SUBSTR[1] JNE NOTEQUAL INC SI MOV AL,[SI] CMP AL,SUBSTR[2] JE EQUAL NOTEQUAL: MOV POS,-1 DISPLAY RTN JMP EXIT EQUAL: MOV DL,POS ADD DL,30H MOV AH,2 INT 21H
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EXIT: MOV AH,4CH INT 21H CODE ENDS END START
Ans.4(b) DATA SEGMENT PACKEDBCD DB 52H BINARY DB ? DATA ENDS CODE SEGMENT ASSUME DS:DATA CS:CODE START: MOV AX,DATA MOV DS,AX MOV AH,0 MOV AL,PACKEDBCD
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MCS-012 SOLVED ASSIGNMENT FOR BCA/MCA MOV BL,AL AND AL,0F0H AND BL,0FH MOV CL,04H ROL AL,CL MOV CL,10 MUL CL ADD AL,BL MOV AH,AL MOV BINARY,AH MOV AH,4CH INT 21H CODE ENDS END START
Screen shots :-
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MCS-012 SOLVED ASSIGNMENT FOR BCA/MCA Before Execution :-
After Execution :-
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Ans.4(c) (i). 8086 has a concept of Memory Segmentation. It is a method where the whole memory is segmented (divided) into smaller parts called segments. These segments are • Code Segment (CS) • Stack Segment (SS) • Data Segment (DS) • Extra Segment (ES) Each Segment has a corresponding correspond ing 16-bit Segment Register which holds the Base Address (starting Address) of the Segment. At any given time, 8086 can address 16-bit x 64KB = 256 KB of memory chunk out of 1MB. 8086 has 20bit address line. So the maximum value of address that can be addressed by 8086 is 2^20 = 1MB. So 8086 can address the locations ranging between 00000 H to FFFFF H. This 1MB memory is divided into 16 logical segments, each with a memory of 64KB.
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MCS-012 SOLVED ASSIGNMENT FOR BCA/MCA To locate any adress in the memory bank, it needs the Physical address of that memory location. It cannot get the 20-bit Physical Ph ysical adress using the 8086 Address Line or 16 -bit Segment Registers alone. In order to access memory location, you cannot pass 20-bit address directly to the processor. You need to tell the 16-bit address with respect to the segment. This 16-bit address with respect to the part (segment of 64KB) of the memory bank is called the offset. So, Physical Address = Base Address + Offset. Suppose the Data Segment holds the Base Aaddress as 1000h and the data you need is present in the 0020h memory location (Offset) of the Data Segment. The calculation of the actual address is done as follows. 1. Left shift the 16-bit address present in the segment register by 4-bits 0001 0000 0000 0000 (0000) 2. Add the 16-bit offset address to this shifted base address 0001 0000 0000 0000 0000 + 0000 0000 0010 0000 -------------------------------------0001 0000 0000 0010 0000
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So the actual address turns out to be 10020h. At any point of time we can c an change the base address of the segment registers and use the memory locations in those segments using the offset.
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