[ANSWERED / SOLVED]GDB Init Question

freddie · Sep 16th 2009, 4:03am

I'm using the J-Link USB tool with an LPC2129. I'm usng the sample generic LPC2xxx_FLASH_JLink.gdb file to initialize the GDB Server (below).

I changed the device to LPC2129.

I'm not sure what the following line is doing. Should I change this to match the memory in the LPC2129?

# Map at the first 64 bytes the internal flash
monitor long 0xE01FC040 = 0x00000001

Brainfuck Source Code

#File : LPC2xxx_FLASH_JLink.gdb
#Purpose: GDB command file for NXP LPC2xxx series +
# J-Link (J-Link GDB Server)
# FLASH is mapped at 0x00000000
#-------- END-OF-HEADER -----------------------------------------------
#
# Connect to the J-Link gdb server
target remote localhost:2331
# Select flash device
monitor flash device = LPC2129
# Enable FlashDL and FlashBPs
monitor flash download = 1
monitor flash breakpoints = 1
# Set gdb server to little endian
monitor endian little
# Set speed to JTAG speed adaptive
monitor speed adaptive
# Reset the target
monitor reset 0
# Map at the first 64 bytes the internal flash
monitor long 0xE01FC040 = 0x00000001
break main
load
continue

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SEGGER - Alex · Sep 16th 2009, 10:51am

Hello freddie,

for clarification what this line does:
>>MEMORY MAPPING CONTROL
>>>>The Memory Mapping Control alters the mapping of the interrupt vectors that appear beginning at address 0x00000000. This
>>allows code running in different memory spaces to have control of the interrupts.
>>Memory Mapping Control Register (MEMMAP - 0xE01FC040)

The memmap register is described in LPC2119/2129/2194/2292/2294 USER MANUAL
in chapter "System control block -> Memory mapping control"

Short: For your case this works.

Best regards
Alex

Brainfuck Source Code

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