What does In [*] at the upper left-hand of the cell mean when running a jupyter notebook and how I can resolve this?
Please any one can help me.
It means that the cell is running, or in queue to be run. You don't solve it really, you just have to wait for the cell to finish executing. If you think the cell has gotten stuck for some reason, you can try to interrupt it by going to "Kernel" in the menu bar and selecting "Interrupt". When the cell is finished running, a number will appear instead of a star. The number is the order in which that cell was executed in this session.
If you have a cell that takes a really long time to execute, I suggest outputting some periodic status to the user. I like to do something like:
update_freq = 100 # or however often you want it to update you
print('Running', end='', flush=True)
for iter in range(a_whole_lot_of_iterations):
# ... code that runs for a long time or for a ton of iterations ...
if iter >= update_freq and iter % summary_freq == 0:
print('.', end='', flush=True) # periodically print something
print('done!')
You can get as fancy as you want with this approach, printing all kinds of cool stuff in your first line as a header, controlling how many updates are printed total to the screen so you can create a sort of ASCII status bar, etc. I found this guide to the Python string formatting language helpful when figuring out how to make nicely formatted status updates.
I am reading sensor output as square wave(0-5 volt) via oscilloscope. Now I want to measure frequency of one period with Beaglebone. So I should measure the time between two rising edges. However, I don't have any experience with working Beaglebone. Can you give some advices or sample codes about measuring time between rising edges?
How deterministic do you need this to be? If you can tolerate some inaccuracy, you can probably do it on the main Linux OS; if you want to be fancy pants, this seems like a potential use case for the BBB's PRU's (which I unfortunately haven't used so take this with substantial amounts of salt). I would expect you'd be able to write PRU code that just sits with an infinite outerloop and then inside that loop, start looping until it sees the pin shows 0, then starts looping until the pin shows 1 (this is the first rising edge), then starts counting until either the pin shows 0 again (this would then be the falling edge) or another loop to the next rising edge... either way, you could take the counter value and you should be able to directly convert that into time (the PRU is states as having fixed frequency for each instruction, and is a 200Mhz (50ns/instruction). Assuming your loop is something like
#starting with pin low
inner loop 1:
registerX = loadPin
increment counter
jump if zero registerX to inner loop 1
# pin is now high
inner loop 2:
registerX = loadPin
increment counter
jump if one registerX to inner loop 2
# pin is now low again
That should take 3 instructions per counter increment, so you can get the time as 3 * counter * 50 ns.
As suggested by Foon in his answer, the PRUs are a good fit for this task (although depending on your requirements it may be fine to use the ARM processor and standard GPIO). Please note that (as far as I know) both the regular GPIOs and the PRU inputs are based on 3.3V logic, and connecting a 5V signal might fry your board! You will need an additional component or circuit to convert from 5V to 3.3V.
I've written a basic example that measures timing between rising edges on the header pin P8.15 for my own purpose of measuring an engine's rpm. If you decide to use it, you should check the timing results against a known reference. It's about right but I haven't checked it carefully at all. It is implemented using PRU assembly and uses the pypruss python module to simplify interfacing.
EDIT: I realized that I, unfortunately, overlooked a semicolon at the end of the while statement in the first example code and misinterpreted it myself. So there is in fact an empty loop for threads with threadIdx.x != s, a convergency point after that loop and a thread waiting at this point for all the others without incrementing the s variable. I am leaving the original (uncorrected) question below for anyone interested in it. Be aware, that there is a semicolon missing at the end of the second line in the first example and thus, s++ has nothing in common with the cycle body.
--
We were studying serialization in our CUDA lesson and our teacher told us that a code like this:
__shared__ int s = 0;
while (s != threadIdx.x)
s++; // serialized code
would end up with a HW deadlock because the nvcc compiler puts a reconvergence point between the while (s != threadIdx.x) and s++ statements. If I understand it correctly, this means that once the reconvergence point is reached by a thread, this thread stops execution and waits for the other threads until they reach the point too. In this example, however, this never happens, because thread #0 enters the body of the while loop, reaches the reconvergence point without incrementing the s variable and other threads get stuck in an endless loop.
A working solution should be the following:
__shared__ int s = 0;
while (s < blockDim.x)
if (threadIdx.x == s)
s++; // serialized code
Here, all threads within a block enter the body of the loop, all evaluate the condition and only thread #0 increments the s variable in the first iteration (and loop goes on).
My question is, why does the second example work if the first hangs? To be more specific, the if statement is just another point of divergence and in terms of the Assembler language should be compiled into the same conditional jump instruction as the condition in the loop. So why isn't there any reconvergence point before s++ in the second example and has it in fact gone immediately after the statement?
In other sources I have only found that a divergent code is computed independently for every branch - e.g. in an if/else statement, first the if branch is computed with all else-branched threads masked within the same warp and then the other threads compute the else branch while the first wait. There's a reconvergence point after the if/else statement. Why then does the first example freeze, not having the loop split into two branches (a true branch for one thread and a waiting false branch for all the others in a warp)?
Thank you.
It does not make sense to put the reconvergence point between the call to while (s != threadIdx.x) and s++;. It disrupts the program flow since the reconvergence point for a piece of code should be reachable by all threads at compile time. Below picture shows the flowchart of your first piece of code and possible and impossible points of reconvergence.
Regarding this answer about recording the convergence point via SSY instruction, I created below simple kernel resembling your first piece of code
__global__ void kernel_1() {
__shared__ int s;
if(threadIdx.x==0)
s = 0;
__syncthreads();
while (s == threadIdx.x)
s++; // serialized code
}
and compiled it for CC=3.5 with -O3. Below is the result of using cuobjdumbinary tool for the output to observe the CUDA assembly. The result is:
I'm not an expert in reading CUDA assembly but I can see while loop condition checks in lines 0038 and 00a0. At line 00a8, it branches to 0x80 if it satisfies the while loop condition and executes the code block again. The introduction of the reconvergence point is at line 0058 introducing line 0xb8 as the reconvergence point which is after the loop condition check near the exit.
Overall, it is not clear what you're trying to achieve with this piece of code. Also in the second piece of code, the reconvergence point should be again after while loop code block (I don't mean between while and if).
The reason why it "hangs" is neither a HW deadlock nor branching, at least not directly. You produce an endless loop for one or multiple threads (as already suspected).
In your example, there isn't really a convergence point. Since you do not use any synchronization, there aren't any threads that actually wait. What happens here with the while-loop is pretty much a busy-wait.
A kernel only finishes if all threads return. Since you have one (or multiple) endless loops (by accident maybe even none - this is unlikely however) the kernel will never finish.
You declared a shared variable s. This variable is known to all threads within a block.
With your while-statement you basically say (to each thread): increment s until it reaches the value of your (local) thread id. Since all threads are incrementing s in parallel, you introduce race conditions.
Example:
List item
Thread 5 is looping and checking for s to become 5
s is 4
Two threads increment s, it becomes 6
At the same time thread 5 only reached the end of its loop.
Now it reaches the next loop iteration and checks for s and it's not 5.
Thread 5 will never be able to finish since you check via == and the value of s already exceeded the value of the thread id.
Also your solution is quite confusing, because each thread executes the serialized code consecutively (which probably was the intention after all - even though that actually is strange):
Thread 0 will execute the serialized code
After that, thread 1 will execute the serialized code
and so on
Most examples show a program where each thread works on some code, then all threads are synchronized and only single thread executes some more code (maybe it needed the results of all threads).
So, your second example "works" because no thread is stuck in an endless loop, however I can't think of a reason why anyone would use such a code,
since it is confusing and, well, not parallel at all.
We can create a real-time monitor for a variable like this:
CreatePalette#Panel#Row[{"x = ", Dynamic[x]}]
(This is more interesting and useful if x happens to be something like $Assumptions. It's so easy to set a value and then forget about it.)
Unfortunately this stops working if the kernel is re-launched (Quit[], then evaluate something). The palette won't show changes in the value of x any more.
Is there a way to do this so it keeps working even across kernel sessions? I find myself restarting the kernel quite often. (If the resulting palette causes the kernel to be automatically started after Quit that's fine.)
Update: As mentioned in the comments, it turns out that the palette ceases working only if we quit by evaluating Quit[]. When using Evaluation -> Quit Kernel -> Local, it will keep working.
Link to same question on MathGroup.
I can only guess, because on my Ubuntu here the situations seems buggy. The trick with the Quit from the menu like Leonid suggested did not work here. Another one is: on a fresh Mathematica session with only one notebook open:
Dynamic[x]
x = 1
Dynamic[x]
x = 2
gives as expected
2
1
2
2
Typing in the next line Quit, evaluating and typing then x=3 updates only the first of the Dynamic[x].
Nevertheless, have you checked the command
Internal`GetTrackedSymbols[]
This gives not only the tracked symbols but additionally some kind of ID where the dynamic content belongs. If you can find out, what exactly these numbers are and investigate in the other functions you find in the Internal context, you may be able to add your palette Dynamic-content manually after restarting the kernel.
I thought I had something like that with
Internal`SetValueTrackExtra
but I'm currently not able to reproduce the behavior.
#halirutan's answer jarred my memory...
Have you ever come across: Experimental/ref/ValueFunction? (documentation address)
Although the documentation contains no examples, the 'more information' section provides the following tidbit:
The assignment ValueFunction[symb] = f specifies that whenever
symb gets a new value val, the expression f[symb,val] should be
evaluated.
GDB has a new version out that supports reverse debug (see http://www.gnu.org/software/gdb/news/reversible.html). I got to wondering how that works.
To get reverse debug to work it seems to me that you need to store the entire machine state including memory for each step. This would make performance incredibly slow, not to mention using a lot of memory. How are these problems solved?
I'm a gdb maintainer and one of the authors of the new reverse debugging. I'd be happy to talk about how it works. As several people have speculated, you need to save enough machine state that you can restore later. There are a number of schemes, one of which is to simply save the registers or memory locations that are modified by each machine instruction. Then, to "undo" that instruction, you just revert the data in those registers or memory locations.
Yes, it is expensive, but modern cpus are so fast that when you are interactive anyway (doing stepping or breakpoints), you don't really notice it that much.
Note that you must not forget the use of simulators, virtual machines, and hardware recorders to implement reverse execution.
Another solution to implement it is to trace execution on physical hardware, such as is done by GreenHills and Lauterbach in their hardware-based debuggers. Based on this fixed trace of the action of each instruction, you can then move to any point in the trace by removing the effects of each instruction in turn. Note that this assumes that you can trace all things that affect the state visible in the debugger.
Another way is to use a checkpoint + re-execution method, which is used by VmWare Workstation 6.5 and Virtutech Simics 3.0 (and later), and which seems to be coming with Visual Studio 2010. Here, you use a virtual machine or a simulator to get a level of indirection on the execution of a system. You regularly dump the entire state to disk or memory, and then rely on the simulator being able to deterministically re-execute the exact same program path.
Simplified, it works like this: say that you are at time T in the execution of a system. To go to time T-1, you pick up some checkpoint from point t < T, and then execute (T-t-1) cycles to end up one cycle before where you were. This can be made to work very well, and apply even for workloads that do disk IO, consist of kernel-level code, and performs device driver work. The key is to have a simulator that contains the entire target system, with all its processors, devices, memories, and IOs. See the gdb mailinglist and the discussion following that on the gdb mailing list for more details. I use this approach myself quite regularly to debug tricky code, especially in device drivers and early OS boots.
Another source of information is a Virtutech white paper on checkpointing (which I wrote, in full disclosure).
During an EclipseCon session we also asked how they do this with the Chronon Debugger for Java. That one does not allow you to actually step back, but can play back a recorded program execution in such a way that it feels like reverse debugging. (The main difference is that you cannot change the running program in the Chronon debugger, while you can do that in most other Java debuggers.)
If I understood it correctly, it manipulates the byte code of the running program, such that every change of an internal state of the program is recorded. External states don't need to be recorded additionally. If they influence your program in some way, then you must have an internal variable matching that external state (and therefore that internal variable is enough).
During playback time they can then basically recreate every state of the running program from the recorded state changes.
Interestingly the state changes are much smaller than one would expect on first look. So if you have a conditional "if" statement, you would think that you need at least one bit to record whether the program took the then- or the else-statement. In many cases you can avoid even that, like in the case that those different branches contain a return value. Then it is enough to record only the return value (which would be needed anyway) and to recalculate the decision about the executed branch from the return value itself.
Although this question is old, most of the answers are too, and as reverse-debugging remains an interesting topic, I'm posting a 2015 answer. Chapters 1 and 2 of my MSc thesis, Combining reverse debugging and live programming towards visual thinking in computer programming, covers some of the historical approaches to reverse debugging (especially focused on the snapshot-(or checkpoint)-and-replay approach), and explains the difference between it and omniscient debugging:
The computer, having forward-executed the program up to some point, should really be able to provide us with information about it. Such an improvement is possible, and is found in what are called omniscient debuggers. They are usually classified as reverse debuggers, although they might more accurately be described as "history logging" debuggers, as they merely record information during execution to view or query later, rather than allow the programmer to actually step backwards in time in an executing program. "Omniscient" comes from the fact that the entire state history of the program, having been recorded, is available to the debugger after execution. There is then no need to rerun the program, and no need for manual code instrumentation.
Software-based omniscient debugging started with the 1969 EXDAMS system where it was called "debug-time history-playback". The GNU debugger, GDB, has supported omniscient debugging since 2009, with its 'process record and replay' feature. TotalView, UndoDB and Chronon appear to be the best omniscient debuggers currently available, but are commercial systems. TOD, for Java, appears to be the best open-source alternative, which makes use of partial deterministic replay, as well as partial trace capturing and a distributed database to enable the recording of the large volumes of information involved.
Debuggers that do not merely allow navigation of a recording, but are actually able to step backwards in execution time, also exist. They can more accurately be described as back-in-time, time-travel, bidirectional or reverse debuggers.
The first such system was the 1981 COPE prototype ...
mozilla rr is a more robust alternative to GDB reverse debugging
https://github.com/mozilla/rr
GDB's built-in record and replay has severe limitations, e.g. no support for AVX instructions: gdb reverse debugging fails with "Process record does not support instruction 0xf0d at address"
Upsides of rr:
much more reliable currently. I have tested it relatively long runs of several complex software.
also offers a GDB interface with gdbserver protocol, making it a great replacement
small performance drop for most programs, I haven't noticed it myself without doing measurements
the generated traces are small on disk because only very few non-deterministic events are recorded, I've never had to worry about their size so far
rr achieves this by first running the program in a way that records what happened on every single non-deterministic event such as a thread switch.
Then during the second replay run, it uses that trace file, which is surprisingly small, to reconstruct exactly what happened on the original non-deterministic run but in a deterministic way, either forwards or backwards.
rr was originally developed by Mozilla to help them reproduce timing bugs that showed up on their nightly testing the following day. But the reverse debugging aspect is also fundamental for when you have a bug that only happens hours inside execution, since you often want to step back to examine what previous state led to the later failure.
The following example showcases some of its features, notably the reverse-next, reverse-step and reverse-continue commands.
Install on Ubuntu 18.04:
sudo apt-get install rr linux-tools-common linux-tools-generic linux-cloud-tools-generic
sudo cpupower frequency-set -g performance
# Overcome "rr needs /proc/sys/kernel/perf_event_paranoid <= 1, but it is 3."
echo 'kernel.perf_event_paranoid=1' | sudo tee -a /etc/sysctl.conf
sudo sysctl -p
Test program:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int f() {
int i;
i = 0;
i = 1;
i = 2;
return i;
}
int main(void) {
int i;
i = 0;
i = 1;
i = 2;
/* Local call. */
f();
printf("i = %d\n", i);
/* Is randomness completely removed?
* Recently fixed: https://github.com/mozilla/rr/issues/2088 */
i = time(NULL);
printf("time(NULL) = %d\n", i);
return EXIT_SUCCESS;
}
compile and run:
gcc -O0 -ggdb3 -o reverse.out -std=c89 -Wextra reverse.c
rr record ./reverse.out
rr replay
Now you are left inside a GDB session, and you can properly reverse debug:
(rr) break main
Breakpoint 1 at 0x55da250e96b0: file a.c, line 16.
(rr) continue
Continuing.
Breakpoint 1, main () at a.c:16
16 i = 0;
(rr) next
17 i = 1;
(rr) print i
$1 = 0
(rr) next
18 i = 2;
(rr) print i
$2 = 1
(rr) reverse-next
17 i = 1;
(rr) print i
$3 = 0
(rr) next
18 i = 2;
(rr) print i
$4 = 1
(rr) next
21 f();
(rr) step
f () at a.c:7
7 i = 0;
(rr) reverse-step
main () at a.c:21
21 f();
(rr) next
23 printf("i = %d\n", i);
(rr) next
i = 2
27 i = time(NULL);
(rr) reverse-next
23 printf("i = %d\n", i);
(rr) next
i = 2
27 i = time(NULL);
(rr) next
28 printf("time(NULL) = %d\n", i);
(rr) print i
$5 = 1509245372
(rr) reverse-next
27 i = time(NULL);
(rr) next
28 printf("time(NULL) = %d\n", i);
(rr) print i
$6 = 1509245372
(rr) reverse-continue
Continuing.
Breakpoint 1, main () at a.c:16
16 i = 0;
When debugging complex software, you will likely run up to a crash point, and then fall inside a deep frame. In that case, don't forget that to reverse-next on higher frames, you must first:
reverse-finish
up to that frame, just doing the usual up is not enough.
The most serious limitations of rr in my opinion are:
https://github.com/mozilla/rr/issues/2089 you have to do a second replay from scratch, which can be costly if the crash you are trying to debug happens, say, hours into execution
https://github.com/mozilla/rr/issues/1373 x86 only
UndoDB is a commercial alternative to rr: https://undo.io Both are trace / replay based, but I'm not sure how they compare in terms of features and performance.
Nathan Fellman wrote:
But does reverse debugging only allow you to roll back next and step commands that you typed, or does it allow you to undo any number of instructions?
You can undo any number of instructions. You're not restricted to, for instance,
only stopping at the points where you stopped when you were going forward. You can
set a new breakpoint and run backwards to it.
For instance, if I set a breakpoint on an instruction and let it run until then, can I then roll back to the previous instruction, even though I skipped over it?
Yes. So long as you turned on recording mode before you ran to the breakpoint.
Here is how another reverse-debugger called ODB works. Extract:
Omniscient Debugging is the idea of
collecting "time stamps" at each
"point of interest" (setting a value,
making a method call,
throwing/catching an exception) in a
program and then allowing the
programmer to use those time stamps to
explore the history of that program
run.
The ODB ... inserts
code into the program's classes as
they are loaded and when the program
runs, the events are recorded.
I'm guessing the gdb one works in the same kind of way.
Reverse debugging means you can run the program backwards, which is very useful to track down the cause of a problem.
You don't need to store the complete machine state for each step, only the changes. It is probably still quite expensive.