"X86 assembly language: From the actual mode to the protection mode" test points and exercise answers

Detection Point 1.1: In Order: 255-56091

Detection Point 1.2: In order respectively: 1000 1010 1100 1111 11001 1000000 1100100 11111111 1111101000 1111111111111111 100000000000000000000< /c7>

Detection Point 1.3: In Order: 8-Ten 1741 1022 4092 65535

detection Point 1.4: In Order: 8 A C F 100000 ff 3e8 FFFF

detection point 1.5:1. by order: 11 1010 1100 1111 100000 111111 1011111110 1111111111111111 100111111100000001011101 1111100110011111 111111011111111

2. in order:1/1 11/3 0101/5 111/7 1001/9 1011/a 1101/d 1111/f 0/0 10/2 100 /4 110/6 1000/8 1100/c 1110/e

detection point 1.6:1. 4092/111111111100 2.27b6100/10011110110110000100000000

The 1th exercise: 1. 5 C =15d=1111b =12d=1100b =10d=1010b =8h=1000b =11d=1011b =14d=1110b =16d=10000b 2.10101 10001111 1000000000 1FF

detection point 2.1:1. (2) (+) (4) (+ 2). (7) (8) The highest bit 3. ( xx) (0F) (8) (00, 02, 04, 06, 08, 0 A, 0C, 0E) double word, 00, 04, 08, 0C

Detection Point 2.2: a3d8h

detection point 2.3:1. 8 (AX BX CX DX SI DI BP SP) (AH AL BH BL CH CL DH DL) 2. ( A) (C) (D F) 3. (A B C D F)

The 2nd exercise: 1. 64 of 2. 25BC0H~35BBFH

detection point 3.1:1. (slightly) 2. (B) ( A) (C)

The 3rd exercise: 1. 00H, 35H, 40H 2. 49H (i.e. 73 bytes)

detection point 4.1:1. (0) (0) (1) (0) (0) (1) 2. (A B C)

detection point 4.2:1. (slightly) 2. (slightly) 3. A, S, m three letters should be displayed in the upper corner of the screen

detection point 5.1:1. ( 0xb8000) (0xb800) (0xf9e) (0x27) (0x48) 2. (E F G H J L) A error is due to an attempt to transmit a 16-bit word to a 8-bit register; The reason for the B error is that the immediate number is transmitted to the segment register; C error is caused by 8-bit register Al to the segment register transmission; The reason for the D error is that the length of the memory operand is not indicated; The reason for the error is that two registers do not match; The reason for the K error is to transfer between two memory units.

Detection Point 5.2:the DB line should not appear 0xf000

Detection Point 6.3:0xf0 0xff 0x81 0xffff 0x8a08

Detection Point 6.4:

1:

When zf= (1), it indicates that the calculated result is zero;    The JZ directive means that when the zf= (1) is calculated, the result is 0 o'clock transfer;    The JE directive means that when the zf= (1) is calculated, the result is 0 o'clock transfer;    The JNZ directive means that when the zf= (0) is not transferred when the result of the calculation is not (0); The JNE directive means that when the zf= (0) is not transferred when the result of the calculation is not (0); 2:

CMP AX,BX

JA LBB

Je lbz

JB LBL

The 1th Chapter Exercises:

Detection Point 12.2: The current stack segment descriptor B bit is 1, the base address is 0x00700000, the threshold value is 0xFFFFE. Then, in 32-bit mode, the valid address range for this stack segment is 0x00700000~ (). When the content of ESP is 0xfffff002, can you still press a double word? Why? as stated in the book, when the B-bit of the stack descriptor is 1 o'clock, the segment bounds are in 4KB units, so the actual segment boundary value used is 0xffffe*0x1000+0xfff=0xffffefff, and then according to

The actual segment bounds +1≤ (the content of the ESP-the length of the operand) ≤0xffffffff

Can get the scope of ESP for 0xffffefff+1<=esp<=0xffffffff, i.e. 0XFFFFF000<=ESP<=0XFFFFFFFF, With the base address of the stack segment given in the title 0x007c0000 plus the minimum and maximum offset allowed by ESP, the lowest end of the stack is 0x007c0000+0xfffff000=0x007bf000 and the highest end address is 0x007c0000+ 0XFFFFFFFF=0X007BFFFF, the valid address range for the stack segment in the topic is 0X007BF000~0X007BFFFF. When the content of ESP is 0xfffff002, the corresponding physical address is 0x007c0000+0xfffff002=0x007bf002, when pressed into a double word, esp-4=0xfffff002-4=0xffffeffe, The corresponding physical address is 0x007c0000+0xffffeffe=0x007beffe, because the address is below the lowest-end address of the stack, so it cannot be pressed into a double-word.

"X86 assembly language: From the actual mode to the protection mode" test points and exercise answers