domingo, 5 de janeiro de 2014

Teste com DEBUG

Teste com DEBUG

Pentium 4

C:\WINDOWS\Desktop>debug
-d 0040:0000 L8
0040:0000  F8 03 F8 02 00 00 00 00                           ........

Note que voce lê de traz para frente, exemplo

03F8 e 02F8 (veja que eu li de 2 em 2 de traz para frente F8 03 eh 03F8)


Compaq
c:\debug
-d 0040:0000 L8

0040:0000 78 03 00 00 00 00 08 02 x.......

0378 E 0208



Analisando a Tabela de Interrupcao do DOS

Exercise. Use debug to examine the interrupt vector table which is assigned to IRQ 0 through IRQ 7.

 
-d 00000:0020 20
 
B3 10 3B 0B 73 2C 3B 0B-57 00 70 03 8B 3B 3B 0B
ED 3B 3B 0B AC 3A 3B 0B-B7 00 70 03 F4 06 70 00

pois
  Pegue a seguencia e separe em grupos de 4 pois a memoria eh segmentada
[B3 10 3B 0B] IRQ 0 0B3B:10B3
[73 2C 3B 0B] IRQ 1
[57 00 70 03] IRQ 2
[8B 3B 3B 0B] IRQ 3
[ED 3B 3B 0B] IRQ 4
[AC 3A 3B 0B] IRQ 5
[B7 00 70 03] IRQ 6
[F4 06 70 00] IRQ 7 0070:06F4

 

(Recall that the table allocation for INT 8 begins at 0x0020).
From this I can see that the address for the interrupt service routine associated with IRQ 0 is 0B3B:10B3. For IRQ 7, 0070:06F4. You should be able to see the algorithm I used to obtain this.
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