22) Explain how storage space allocated for initialized data by the following in

Question

22) Explain how storage space allocated for initialized data by the following instructions TSU ENGINEERING DB Y NUMBER DW 45066 NEG NUMBER DW -45066 23) All computers use the addition process to implement subtraction (TRUE/FALSE)? 24) Computers use Adders Circuitry in conjunction with 2's complement circuitry to perform subtraction 25) Write an assembly language instructions to store 25H and 5H in microprocessor registers, and then: () Multiply 25H by 5H, and(ii Divide 25H by 5H 26) Given the logical address CS:IP (2550-9FS5H), compute the physical address of this memory location. 27) IfCS-24F6H and IP 634AH, show (a) the logical address, and (b) the offset address. Calculate (c) the physical address, (d) the lower range, and (e) the upper range of the code segment.

Explanation / Answer

22.

Following are the points to take in mind while allocating

Let say starting byte is 1001

TSU_ENGINEERING DB 'Y' // ASCII of y = 79 H , this will assign 1 byte to it i.e. 1001

NUMBER DW 45066 // 45066D = B00A H , this will assign 2 byte i.e. 1002-1003

NEG_NUMBER DW -45066 // -45066D = 4FF6 H (2 COMPLIMENT OF -45066d = 20470D) ,this will assign 2 byte i.e. 1004-1005

so DB - 1001

DW - 1002-1003

DW - 1004-1005

Each byte of character is stored as its ASCII value in hexadecimal.

Each decimal value is automatically converted to its 16-bit binary equivalent and stored as a hexadecimal number.

Negative numbers are converted to its 2's complement representation.

Short and long floating-point numbers are represented using 32 or 64 bits, respectively.

DB Define Byte allocates 1 byte DW Define Word allocates 2 bytes

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.