第一个串口这个Active下的04是由什么决定的?
Try to open Drivers\Active\04
红外串口Active下的05是由什么决定的?
Try to open Drivers\Active\05
下面是他们的注册表
;=============== UART0 (physical COM1 connector P1) (Serial) ===============
[HKEY_LOCAL_MACHINE\Drivers\BuiltIn\Serial]
"DeviceArrayIndex"=dword:0
"Irq"=dword:1c
"MemBase"=dword:50000000
"MemLen"=dword:2C
"InterruptBitsShift"=dword:0 ; UART 0 Interrupt Sub Register shift bit.
"ISTTimeouts"=dword:200 ; every 512 ticks checking Modem status.
"Prefix"="COM"
"Index"=dword:1
"Dll"="serial_smdk2440.dll"
"Order"=dword:0
"Priority"=dword:0
"Port"="COM1:"
"DeviceType"=dword:0
"FriendlyName"=LOC_DISPLAYNAME_COM1
"Tsp"="unimodem.dll"
"IClass"="{CC5195AC-BA49-48a0-BE17-DF6D1B0173DD}"
"DevConfig"=hex: 10,00, 00,00, 05,00,00,00, 10,01,00,00, 00,4B,00,00, 00,00, 08, 00, 00, 00,00,00,00
ENDIF
ENDIF
IF BSP_NOIRDA !
; S2440 IrDA(UART2)
[HKEY_LOCAL_MACHINE\Drivers\BuiltIn\IRDA2440]
"DeviceArrayIndex"=dword:1
"Irq"=dword:f
"MemBase"=dword:50008000
"MemLen"=dword:2C
"InterruptBitsShift"=dword:6 ; UART 2 Interrupt Sub Register shift bit.
;"ISTTimeouts"=dword:200 ; every 512 ticks checking Modem status.
"Prefix"="COM"
"Index"=dword:4
"Dll"="serial_smdk2440.Dll"
"Order"=dword:0
"Priority"=dword:0
"Port"="COM4:"
"DeviceType"=dword:0 ; IRDA modem, 0 -> null modem
"FriendlyName"=LOC_DISPLAYNAME_RAWIR
"IClass"="{CC5195AC-BA49-48a0-BE17-DF6D1B0173DD}"
"DevConfig"=hex: 10,00, 00,00, 05,00,00,00, 10,01,00,00, 00,4B,00,00, 00,00, 08, 00, 00, 00,00,00,00
ENDIF