Line 14 introduces the instruction mov. It "moves" meaning updates the content of al segment with a hex value 22h which is a byte. Then the program calls again DumpRegs. The EAX register is updated in the least significant byte. Insert a comment here in your code and explain why the EAX register shows 22h in the last byte, when you updated the al register segment. Make sure you also explain why the rest of the bytes in EAX are unaffected.

Line 16 does the same, only that it updates the ax segment. When DumpRegs is called, you will see EAX with a certain value. Explain in your code, in a comment, why EAX shows that value. Make sure you explain why the upper two bytes of EAX are not affected.

In line 18, we move 33h hex value, this time into ah. Explain the result in a comment in your code.

In line 20, we move 4444h hex value, this time into eax. Explain the result in a comment in your code. Make sure you explain why the upper two bytes of EAX were cleared although the program updated just the last two bytes.

14 10 .code 11 main PROC 12 call clrscr 131 call DumpRegs mov al, 22h 15 call DumpRegs mov ax, 2222h call DumpRegs mov ah, 33h call DumpRegs 201 mov eax, 4444h call DumpRegs 19 exit main ENDP 24 25 26 END main