Question

Problem 4. [10 pts] Find the result and resulting NZCV flags for each 16-bit operation. operation result 0x8004 0x8005 0x8000 0x6000 0xA000 0x6000 0x3000 - 0xA000 Problem 5. [5 pts] Explain the concept of endinanness and give examples Problem 6. [S pts] What is the purpose of a breakpoint and explain different types (with examples if necessary)

Answer 1

Answer 4)

0x8004 - 0x8005 = 0xFFFF (-1 in 2's complement format)

Therefore N = 1 (as the result is negative)

Z = 0 (as the result is non-zero)

C = 0 (as the result fits in 16 bits)

V = 0 (there is no overflow)

Please note that overflow bit is set when addition of two positive (P) number gives a negative number or addition of two negative (N) numbers gives a positive number

Also in case of subtraction

if P - N = N then there is overflow

or if N - P = P then also there is overflow

Likewise

0x8000 + 0x6000 = 0xE000

N = 1 (result is negative)

Z = 0

C = 0

V = 0 as this is case of adding a negative number 0x8000 with a positive number 0X6000

Next

0xA000 + 0x6000 = 0x10000

N = 0 (the result in 0 which non-negative)

Z = 1 as the lower sixteen bits have become 0

C = 1 as the result of adding two 16-bit numbers in this case cannot be contained in 16 bits.

V = 0, there is no overflow here

Next

0x3000 - 0xA000 = 0x9000

N = 1, the result is negative

Z = 0

C = 1 ( a borrow is generated)

V = 1 (this is a case of Positive - Negative subtraction, and the result is negative. This indicates that overflow has taken place)

Answer 5)

Endian-ness refers to how we store the bytes of a multibyte number in memory. Endian-ness comes in two flavours - Little Endian and Big Endian. I think an example will help you understand the difference immediately:

Suppose we wish too store a 16-bit number 0x1234 in the memory, starting at address 0x00

In Little Endian system this storage will be as shown below:

Address Content

00 Hex 34 Hex

01 Hex 12 Hex

So in this case the lower address stores the lower order byte of the data

In Big Endian system the same number would be stored as

Address Content

00 Hex 12 Hex

01 Hex 34 Hex

So here the higher order byte is stored at the lower address

Answer 6

The term breakpoint generally refers to an address (or addresses) in a program where programmer wishes the program execution to halt temporarily. This is typically done when debugging a program. When the execution is halted at a particular location to let the programmer can examine the contents of various registers, I/O port and memory location to check if these are as expected. This, as you can imagine, would not be possible if program execution was happening at full-speed. These type of breakpoints are called Instruction Breakpoint.

For example consider the following set of instructions:

Instruction 1

Instruction 2

Instruction 3

Instruction 4

Instruction 5

................

Here the programmer has set the breakpoint at Instruction 3. So the execution halts just short of executing this instruction. After that the programmer can examine the registers etc and/or execute instruction 4, 5 one by one (i.e. in single-step mode).

There can also be data breakpoint. In this case the program execution stops when value of a variable or contents of a register /Data memory location acquire a certain value.

Hope this helps.