Practice Problems

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Exercises 13.6 Practice Problems

1.

How many bytes in memory can be accessed if 6 bits are used in a 16-bit instruction for the memory address?

Answer.

64

Solution.

\(2^6 = 64\)

2.

In indirect addressing, a 16-bit pointer register is used to address memory. How many bytes of memory space can be accessed in this scenario?

Answer.

64k

Solution.

\(2^{16} = 65535 = 64\textrm{k}\)

For each of the following questions, use Table 13.6.1 and Y = 0x0103 to determine the value stored in the destination register, as well as the value of pointer register Y, after each of the following subsequent operations. Assume that each of the following instructions are executed in order and will affect the value of Y in subsequent questions.

Table 13.6.1. Data stored in different SRAM addresses.

Address	SRAM Contents
`0x0100`	`0xC1`
`0x0101`	`0xFB`
`0x0102`	`0x3F`
`0x0103`	`0xE9`
`0x0104`	`0x95`
`0x0105`	`0x9A`
`0x0106`	`0x1C`
`0x0107`	`0xDC`

3.

LD r4, Y

Answer.

r4 = 0xE9, Y = 0x0103

4.

LD r5, Y+

Answer.

r5 = 0xE9, Y = 0x0104

5.

LD r6, Y

Answer.

r6 = 0x95, Y = 0x0104

6.

LD r7, Y+3

Answer.

r7 = 0xDC, Y = 0x0107

7.

LD r8, Y-

Answer.

r8 = 0x1C, Y = 0x0106

8.

LDS r9, 0x0102

Answer.

r9 = 0x3F, Y = 0x0106