I use ARM9
I had tried not stepping thru also.
What does coelessing mean ?
thanks
Reply by Will●April 12, 20052005-04-12
shankar.vk@gmail.com wrote in
news:1113222190.466570.106750@l41g2000cwc.googlegroups.com:
> As I said, the memory access works through jtag debugger
> Trace32.
>
> What is the difference in execution flow of memory access
> through debugger
> command and software instruction ? This is the key to
> solve this problem.
stepping the code will cause a drain of the write buffer on
every instruction to syncronize 'memory', Your first post has
expired here - so I don't know what core you are using. Is the
write buffer coelessing?
Reply by ●April 11, 20052005-04-11
As I said, the memory access works through jtag debugger Trace32.
What is the difference in execution flow of memory access through
debugger
command and software instruction ? This is the key to solve this
problem.
Reply by ●April 11, 20052005-04-11
Let me add that there are other memory mapped registers on the fpga
.
I have a problem only accessing the I2C related register.
In terms of execution flow, what is the difference between memory
access through :
a) Trace32 JTAG debugger
and
b) code running on arm ?
Reply by ●April 11, 20052005-04-11
Thanks to all who replied.
I forgot to add that the same C code works when executed from an ARM
core which
is part of the same fpga which contains the I2C register
So the code is perfect.
We analysed the memory access signals such as chip select, write
enable, wait etc
using a TLA . They look OK.
The wait signal comes as an input to the memory controller peripheral
of ARM from the fpga.
thanks
Reply by Thad Smith●April 4, 20052005-04-04
shankar.vk@gmail.com wrote:
> When I try to write into this register from ARM software as
> shown below, write is not happening. ARM is running at 48 Mhz.
>
> int* reg = 0x44400030;
> *reg = 0x3;
>
> The above code is not able to modify the register.
I recommend that you look at the compiler output, either through an
assembly listing or, better yet, through the disassembly output of a
debugger. See what code is being generated. Compare that to what you
know works. This is a powerful technique that all programmers, but
especially embedded systems programmers and systems programmers can
benefit from.
It might have something to do with memory mapping or the width of the
write, but those are just guesses.
Thad
Reply by Jet Morgan●April 3, 20052005-04-03
<shankar.vk@gmail.com> wrote in message
news:1112358355.997835.104340@z14g2000cwz.googlegroups.com...
> int* reg = 0x44400030;
> *reg = 0x3;
>
> The above code is not able to modify the register.
Have you previously written the same value to that
location ? If so, it's possible that your compiler
spots the duplication and doesn't do it a second time
(unless you write something else in between).
This is because "int" refers to a "variable" and
not a memory location, per se. The difference is
that writing to a memory location can have a side
effect. This is what you want, so you may need to
stick a "volatile" qualifier in, to make sure the
compiler really does do a write.
Richard [in PE12]
Reply by RS●April 1, 20052005-04-01
Or this:
int* reg = (int*)0x44400030;
*reg = 0x3L;
Regis (regis@dacafe.com)
"RS" <regis@dacafe.com> wrote in message
news:CVd3e.43674$AN1.42984@fed1read03...
> Try this:
>
> int* reg = (int*)0x44400030;
> *reg = 0x3;
>
> Regis.
>
> <shankar.vk@gmail.com> wrote in message
> news:1112358355.997835.104340@z14g2000cwz.googlegroups.com...
>>
>> Hi,
>>
>> I have an fpga which is accessed from ARM as a
>> memory-mapped device.
>>
>> A register in fpga is used to drive a I2C bus connected to
>> a device . Two bits in the register represent clock and
>> data lines of the bus.
>>
>> When I try to write into this register from ARM software as
>> shown below, write is not happening. ARM is running at 48 Mhz.
>>
>> int* reg = 0x44400030;
>> *reg = 0x3;
>>
>> The above code is not able to modify the register.
>>
>> But when I write using Lauterbach trace32 command as shown below, it
>> works.
>>
>> data.set SD:0x44400030 %LE %LONG 0x3
>>
>> What could be the reason ?
>>
>> Please advise.
>> thanks
>> shankar
>>
>
>
Reply by RS●April 1, 20052005-04-01
Try this:
int* reg = (int*)0x44400030;
*reg = 0x3;
Regis.
<shankar.vk@gmail.com> wrote in message
news:1112358355.997835.104340@z14g2000cwz.googlegroups.com...
>
> Hi,
>
> I have an fpga which is accessed from ARM as a
> memory-mapped device.
>
> A register in fpga is used to drive a I2C bus connected to
> a device . Two bits in the register represent clock and
> data lines of the bus.
>
> When I try to write into this register from ARM software as
> shown below, write is not happening. ARM is running at 48 Mhz.
>
> int* reg = 0x44400030;
> *reg = 0x3;
>
> The above code is not able to modify the register.
>
> But when I write using Lauterbach trace32 command as shown below, it
> works.
>
> data.set SD:0x44400030 %LE %LONG 0x3
>
> What could be the reason ?
>
> Please advise.
> thanks
> shankar
>
Reply by Vadim Borshchev●April 1, 20052005-04-01
On 1 Apr 2005 04:25:56 -0800, <shankar.vk@gmail.com> wrote:
> int* reg = 0x44400030;
> *reg = 0x3;
> [...]
>
> What could be the reason ?