Some simple questions, compilation masters

The register is 32 bits, four bytes of course.

Instructions can be pressed into all 32-bit general-purpose registers in the following order in the stack: EAX, ECX, EDX, EBX, ESP raw values, EBP, ESI, EDI.

3.A total of 8 32-bit general-purpose registers are stacked after the pushad command is executed, so the executed ESP=0013FFC4-4*8=0013FFa4.

Reference: Intel Assembly Language Programming, Fourth Edition, p. 116.

The answer to the word, 1l has been given.

Probably you need an official Intel reference:

mov al，78h ；The immediate number of 78h is transmitted to the accumulator ax of the low-bit byte AL

mov bl，0c3h ；The immediate number 0c3h is passed to the low-bit byte bl of register bx

add al，bl ;Addition without carry The contents of al and bl are added and the result is sent to al for storage The operation affects the state value of the corresponding bit in the flag register.

The addition of hexadecimal numbers 78h+0c3h results in 3bh and the addition of the low digits is 8+3=11, that is, the hexadecimal b has no carry, the high digits are added 7+12 is 19, and every hexadecimal one will produce a carry 1 and the remaining number is 3

Since the highest sign bit = 0 is a positive number in 78h and the highest sign bit = 1 in 78h is a negative number, when added, there will be no overflow, and the two will only get smaller and smaller.

Results: al=3bh, carry flag cf=1, overflow flag of= 0.

Haha, okay, don't forget to give some extra points.

The answer given in the question is wrong.

The correct answer to this question should be: 02ffh

In the problem, the four 16-digit numbers defined by the array, written in hexadecimal form, are 0ffffh (the complement representation of 1), 0002h, 0fffdh (the complement representation of 3), 0004h.

array + 0 : ff

array + 1 : ff

array + 2 : 02

array + 3 : 00

array + 4 : fd

array + 5 : ff

array + 6 : 04

array + 7 : 00

Note that each word occupies two bytes, the low address is the low byte, and the high address is the high byte) instruction mov ax, array+1 is to take a word at array+1 and transfer it to ax.

This word consists of 2 bytes, a byte ff at array+1 is a low-level byte, and a byte 02 at array+2 is a high-level byte.

So, the word taken is 02ffh.

Each machine has a different answer, the answer is correct, but the machine is different! So you'd better test it yourself

1DX is shifted right 8 times, which is equivalent to high and low byte swapping, so it is 2817h.

Then with 0ffh, it is equivalent to taking the low byte, that is, dx=0017h and then comparing it with 17h, equal, so zf=1.

So the answer to question 1: (DX)=0017H, ZF=12 and 0FH XOR, which is equivalent to the high 4 bits unchanged, and the low 4 bits are negated, so al=5ch and 0FH are in place, which is equivalent to the high 4 bits clear 0, and the low 4 bits are unchanged, so BL=03H and 0FH are in place or, which is equivalent to the high 4 bits unchanged, and the low 4 bits all become 1, so cl=5fh

<>ax=3000h 。The following is the result of the verification of Yukong Pei in debug.