1,Check the best answer. This register is the program counter for the CPU we are

Question

1,Check the best answer.

This register is the program counter for the CPU we are studying.

EDI ; ESI ; EFLAGS ; EIP ; EDX

2,Fill in multiple blanks.

Suppose that initially EDX = 0000FFFFh and DX is incremented (DX <-- DX + 1), then EDX = ________h, DX=________h,and DL=________h.

3,Check all that apply.

These registers are accessible 16-bit registers for the CPU we are studying.

DS; CX; AH; DX; SS; EAX; ESI

4,Check all that apply.

For the CPU we are studying, check all registers that are strictly addressing registers: par convention, these registers are meant to store/compute addresses.

ESP; EBX; ESI; CX; EDX; EDI

5,Check the best answer.

Consider the following instruction

MOV EBX, 1A2BC3D4h

MOV AX, BX

After this program is executed, the register AH will contain _____________

C3h; C3D4h; 1A2Bh; D4h; Undetermined

Undetermined ; 2Bh ; C3h ;D4h ;1AH

6,Check the best answer.

Consider the following program

MOV EBX, 1A2BC3D4h

MOV AX, BX

After this program is executed, the register AX will contain ____________; BX will contain _____________.

Explanation / Answer

1. The program counter in this CPU is EIP (Instruction Pointer)

2. EDX = 0000FFFFh which is a 32bit register. whereas DX is 16 bit rigister so DX = FFFFh

Whereas DL and DH are 8 bit registers .. DL is lower 8 bits(LSB) so DL = FFh

3. CX and DX are 16 bit registers. and there are segment registers SS and DS which are also 16 bit.

4.  addressing registers are ESP; EBX; ESI; EDI

5 . EBX = 1A2BC3D4h So BX = C3D4h

MOV AX, BX which is AX= BX finally AX = C3D4h

So AH is (MSB) higher 8 bits which C3 so AH = C3h

6. EBX = 1A2BC3D4h

BX = C3D4h and after MOV AX,BX

AX = C3D4h and BX = C3D4h

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