questions and curiosity

34 branch prefect

32: copy from user

MOVS

Moves the byte, word, or doubleword specified with the second operand (source operand) to the location specified 
with the first operand (destination operand). Both the source and destination operands are located in memory. The 
address of the source operand is read from the DS:ESI or the DS:SI registers (depending on the address-size 
attribute of the instruction, 32 or 16, respectively). The address of the destination operand is read from the ES:EDI 
or the ES:DI registers (again depending on the address-size attribute of the instruction). The DS segment may be 
overridden with a segment override prefix, but the ES segment cannot be overridden.
At the assembly-code level, two forms of this instruction are allowed: the “explicit-operands” form and the “no-
operands” form. The explicit-operands form (specified with the MOVS mnemonic) allows the source and destination 
operands to be specified explicitly. Here, the source and destination operands should be symbols that indicate the 
size and location of the source value and the destination, respectively. This explicit-operands form is provided to 
allow documentation; however, note that the documentation provided by this form can be misleading. That is, the 
source and destination operand symbols must specify the correct type (size) of the operands (bytes, words, or 
doublewords), but they do not have to specify the correct location. The locations of the source and destination 
operands are always specified by the DS:(E)SI and ES:(E)DI registers, which must be loaded correctly before the 
move string instruction is executed. 
The no-operands form provides “short forms” of the byte, word, and doubleword versions of the MOVS instruc-
tions. Here also DS:(E)SI and ES:(E)DI are assumed to be the source and destination operands, respectively. The 
size of the source and destination operands is selected with the mnemonic: MOVSB (byte move), MOVSW (word 
move), or MOVSD (doubleword move).
After the move operation, the (E)SI and (E)DI registers are incremented or decremented automatically according 
to the setting of the DF flag in the EFLAGS register. (If the DF flag is 0, the (E)SI and (E)DI register are incre-mented; 

if the DF flag is 1, the (E)SI and (E)DI registers are decremented.) The registers are incremented or 

decremented by 1 for byte operations, by 2 for word operations, or by 4 for doubleword operations.

Operation
DEST ← SRC;
Non-64-bit Mode:
IF (Byte move)
THEN IF DF = 0
THEN 
(E)SI ← (E)SI + 1; 
(E)DI ← (E)DI + 1; 
ELSE 
(E)SI ← (E)SI – 1; 
(E)DI ← (E)DI – 1; 
FI;
ELSE IF (Word move)
THEN IF DF = 0
(E)SI ← (E)SI + 2; 
(E)DI ← (E)DI + 2; 
FI;
ELSE 
(E)SI ← (E)SI – 2; 
(E)DI ← (E)DI – 2; 
FI;
ELSE IF (Doubleword move)
THEN IF DF = 0
(E)SI ← (E)SI + 4; 
(E)DI ← (E)DI + 4; 
FI;
ELSE 
(E)SI ← (E)SI – 4; 
(E)DI ← (E)DI – 4; 
FI;
FI;
 

Operation
IF AddressSize = 16
    THEN
        Use CX for CountReg;
        Implicit Source/Dest operand for memory use of SI/DI;
    ELSE IF AddressSize = 64
        THEN Use RCX for CountReg; 
        Implicit Source/Dest operand for memory use of RSI/RDI;
    ELSE
        Use ECX for CountReg;
        Implicit Source/Dest operand for memory use of ESI/EDI;
FI;
WHILE CountReg ≠ 0
DO
Service pending interrupts (if any);
Execute associated string instruction;
CountReg ← (CountReg – 1);
IF CountReg = 0
THEN exit WHILE loop; FI;
IF (Repeat prefix is REPZ or REPE) and (ZF = 0)
or (Repeat prefix is REPNZ or REPNE) and (ZF = 1)
THEN exit WHILE loop; FI;
OD;
 

31  64 Bit memory

64 Bit memory

others 

30 size of PIPE

29 Physical address map

28 IOremap

IO shared memory ==> IOremap ==> vmalloc

27: Per CPU

26: 各種鎖

25: smp中

processor A 訪問共享資料Data, A對Data加鎖，此時processor B 也必須要用到Data，怎麼辦？

23: initrd and vmlinux compiling ?

the procedure of buidling initrd and vmlinx 

and the script 

are they compiled with the same source files ?

22: linux kernel startup

grub and linux,  how does it work ? why it need grub ? 

4： float

最大值就是：0 11111110 11111111111111111111111

0 1111 1110 1.111 1111 1111 1111 1111 1111 -- 有效正max
1 1111 1110 1.111 1111 1111 1111 1111 1111 -- 有效負max


V = (-1)^0 * 2^(254-127) * 1.111 1111 1111 1111 1111 1111
  = 2^(127-23) * 1111 1111 1111 1111 1111 1111.
  = 2^104 * 16777215
  = 3.4028234663852885981170418348452e+38


0 0000 0000 0.000 0000 0000 0000 0000 0001 -- 正min (+0)
1 0000 0000 0.000 0000 0000 0000 0000 0001 -- 負min (-0)


+V =  (-1)^0 * 2^(-126) ^ 0.000 0000 0000 0000 0000 0001
     =  + 2^(-149)
     =  + 1.4012984643248170709237295832899e-45
-V =  - 2^(-149)

0 1111 1111 0000 0000 0000 0000 0000 000 -- 正無窮
1 1111 1111 0000 0000 0000 0000 0000 000 -- 負無窮

+V = 正無窮 = 2^(128) =     3.4028236692093846346337460743177e+38
-V = 負無窮 = -2^(128) =  -3.4028236692093846346337460743177e+38

0 1111 1111 xxxx .... -- NaN (not a number)
1 1111 1111 xxxx .... -- NaN

5: pragma

6:  preprocessor output

8: fork exit wait4

task_struct

thread_info

9: spin lock and semaphore

 10: msi

11: stack

12： 時鐘中斷處理函式process

13: driver

14: 理解linux OS 實現中的抽象層次機制

介面層次，oop程式設計方式

Linux 時鐘處理機制

15: CFS

16: interview 

17:  OS course

http://www.cse421.net/lecture/ 

18: kernel ring buffer and printk