NetBurner Community Forum

I am wondering is anyone has experience getting a Nano to run as a ROS (Robot Operating System) ( http://www.ros.org/ ) node using tcp or udp transport? ROS is a publish /subscribe protocol for sharing data across many platforms, sort of like MQTT, but with TONS of tools for modeling, simulating and...

Have not used either, but have similar interest and found:
http://www.7-zip.org/sdk.html and
http://tukaani.org/xz/

My 1st guess was that the format string you're using may include a format specifier parameter that may be inappropriate for the type you're trying to format. However f_getfreespace() just returns and int. Looking at C:\nburn\docs\EFFS\fat_331.pdf the FAT Implementation guide, section 4.18 on page 49...

See:

http://www.wvshare.com/product/PCF8563-RTC-Board.htm
http://www.aliexpress.com/store/product ... 02369.html

Hi Guys: I tried the C:\nburn\examples\NANO54415\OnChipRtc\main.cpp demo.
I found no connection to the 207.200.81.113 NTP server.
I changed it to: 198.123.30.132, or to "164.67.62.194" and both worked.

You can poll the pin before setting to IRQ mode. Once in IRQ mode, you need to reconfigure to GPIO to poll it.

I don't think you need to check the levels or do anything before arming the IRQ. Just call SetPinIrq(); the logic works in either of the two cases: Case 1) Say level is lo when you arm IRQ to act on a falling edge. Nothing happens when you execute SetPinIrq() because even though the level is low, th...

Forgot to add: Once you arm the IRQ to act on a particular edge, there is no need to poll the line: you WILL miss the edge that caused the IRQ!

The default pin functions should be GPIO and so level could be determined before configuring the IRQ. However there may be no need to do that. When you call SetPinIrq() you spec the IRQ polarity to be rising, falling or both edges. So if you spec'd falling edge, and the line was already low, no IRQ ...

Nope, one task per prio! After an IRQ occurs, or when the task switcher runs, the task with the highest priority that's ready to run will run. Read that statement again! When a task is blocked, (ie waiting for a resource, something like a semaphore) it is not ready to run. It becomes ready to run wh...