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Re: RedBoot Question
- From: "Simon" <simoncc at ms46 dot url dot com dot tw>
- To: "HuangQiang" <jameshq at liverpool dot ac dot uk>,<ecos-discuss at sources dot redhat dot com>
- Date: Thu, 22 Nov 2001 18:24:01 +0800
- Subject: Re: [ECOS] RedBoot Question
- References: <KIEBICHBADHFCLGCKOPDOEIHCJAA.jameshq@liv.ac.uk>
Huang,
1. the answer is 512K byte. 0x10000 = 64K such that 0x80000 = 512K
2. you can use ICE to dump the memory 0x0 and 0x1800000 and you will
see the same contents. please have a look at Samsung's document before
asking
questions.
"Not hard but easy"
Simon
----- Original Message -----
From: "HuangQiang" <jameshq@liverpool.ac.uk>
To: "eCos" <ecos-discuss@sources.redhat.com>
Sent: Thursday, November 22, 2001 5:39 PM
Subject: [ECOS] RedBoot Question
Dear All:
In ARM E7T RedBoot User Manual:
Physical Address Range C B Description
0x00000000 - 0x0007ffff Y N SDRAM
0x03f f 0000 - 0x03f f ffff N N Microcontroller registers
0x01820000 - 0x0187ffff N N System flash (mirrored)
In the ARM E7T board there are 512KB SRAM, 512KB Flash memory. compared
with the above redboot memory setting confused me.
Q1: As indicated above in RedBoot: 0x00000000 - 0x0007ffff SRAM ( ?=
(7ffff/1024) KB SRAM (am I right?) ) But where is the on board 512KB SRAM?
(or maybe I made the wrong way)
Q2: What is the meaning of "mirrored" in "System flash(mirrored)"?
Thanks a lot.
Best regards!
Mr Huang Qiang
Postgraduate Student
Room318 B-Block
Department Of Electrical Engineering And Electronics
University Of Liverpool
Brownlow Hill
Liverpool L69,3GJ
UK
Personal website at: http://www.liv.ac.uk/~jameshq