Following my previous question : Unable to implement MPI_Intercomm_create
The problem of MPI_INTERCOMM_CREATE has been solved. But when I try to implement a basic send receive operations between process 0 of color 0 (globally rank = 0) and process 0 of color 1 (ie globally rank = 2), the code just hangs up after printing received buffer.
the code:
program hello
include 'mpif.h'
implicit none
integer tag,ierr,rank,numtasks,color,new_comm,inter1,inter2
integer sendbuf,recvbuf,tag,stat(MPI_STATUS_SIZE)
tag = 22
sendbuf = 222
call MPI_Init(ierr)
call MPI_COMM_RANK(MPI_COMM_WORLD,rank,ierr)
call MPI_COMM_SIZE(MPI_COMM_WORLD,numtasks,ierr)
if (rank < 2) then
color = 0
else
color = 1
end if
call MPI_COMM_SPLIT(MPI_COMM_WORLD,color,rank,new_comm,ierr)
if (color .eq. 0) then
if (rank == 0) print*,' 0 here'
call MPI_INTERCOMM_CREATE(new_comm,0,MPI_Comm_world,2,tag,inter1,ierr)
call mpi_send(sendbuf,1,MPI_INT,2,tag,inter1,ierr)
!local_comm,local leader,peer_comm,remote leader,tag,new,ierr
else if(color .eq. 1) then
if(rank ==2) print*,' 2 here'
call MPI_INTERCOMM_CREATE(new_comm,2,MPI_COMM_WORLD,0,tag,inter2,ierr)
call mpi_recv(recvbuf,1,MPI_INT,0,tag,inter2,stat,ierr)
print*,recvbuf
end if
end
The communication with intercommunication is not well understood by most users, and examples are not as many as examples for other MPI operations. You can find a good explanation by following this link.
Now, there are two things to remember:
1) Communication in an inter communicator always go from one group to the other group. When sending, the rank of the destination is its the local rank in the remote group communicator. When receiving, the rank of the sender is its local rank in the remote group communicator.
2) Point to point communication (MPI_send and MPI_recv family) is between one sender and one receiver. In your case, everyone in color 0 is sending and everyone in color 1 is receiving, however, if I understood your problem, you want the process 0 of color 0 to send something to the process 0 of color 1.
The sending code should be something like this:
call MPI_COMM_RANK(inter1,irank,ierr)
if(irank==0)then
call mpi_send(sendbuf,1,MPI_INT,0,tag,inter1,ierr)
end if
The receiving code should look like:
call MPI_COMM_RANK(inter2,irank,ierr)
if(irank==0)then
call mpi_recv(recvbuf,1,MPI_INT,0,tag,inter2,stat,ierr)
print*,'rec buff = ', recvbuf
end if
In the sample code, there is a new variable irank that I use to query the rank of each process in the inter-communicator; that is the rank of the process in his local communicator. So you will have two process of rank 0, one for each group, and so on.
It is important to emphasize what other commentators of your post are saying: when building a program in those modern days, use moderns constructs like use mpi instead of include 'mpif.h' see comment from Vladimir F. Another advise from your previous question was yo use rank 0 as remote leader in both case. If I combine those 2 ideas, your program can look like:
program hello
use mpi !instead of include 'mpif.h'
implicit none
integer :: tag,ierr,rank,numtasks,color,new_comm,inter1,inter2
integer :: sendbuf,recvbuf,stat(MPI_STATUS_SIZE)
integer :: irank
!
tag = 22
sendbuf = 222
!
call MPI_Init(ierr)
call MPI_COMM_RANK(MPI_COMM_WORLD,rank,ierr)
call MPI_COMM_SIZE(MPI_COMM_WORLD,numtasks,ierr)
!
if (rank < 2) then
color = 0
else
color = 1
end if
!
call MPI_COMM_SPLIT(MPI_COMM_WORLD,color,rank,new_comm,ierr)
!
if (color .eq. 0) then
call MPI_INTERCOMM_CREATE(new_comm,0,MPI_Comm_world,2,tag,inter1,ierr)
!
call MPI_COMM_RANK(inter1,irank,ierr)
if(irank==0)then
call mpi_send(sendbuf,1,MPI_INT,0,tag,inter1,ierr)
end if
!
else if(color .eq. 1) then
call MPI_INTERCOMM_CREATE(new_comm,0,MPI_COMM_WORLD,0,tag,inter2,ierr)
call MPI_COMM_RANK(inter2,irank,ierr)
if(irank==0)then
call mpi_recv(recvbuf,1,MPI_INT,0,tag,inter2,stat,ierr)
if(ierr/=MPI_SUCCESS)print*,'Error in rec '
print*,'rec buff = ', recvbuf
end if
end if
!
call MPI_finalize(ierr)
end program h
Related
In my experience Matlab performs publish subscribe operations with ROS slow for some reason. I work with components as defined in an object class as shown below, where I made a test-class. Normally objects of comparable structure are used to control mobile robots.
To quantify performance tested required time for an operation and got the following results:
1x publishing a message + 1x simple subscriber callback : 3.7ms
Simply counting in a callback (per count): 2.1318e-03 ms
Creating a new message with msg1 = rosmessage(obj.publisher) adds 3.6-4.3ms per iteration
Pinging myself indicated communication latency of 0.05 ms
The times required for a simple publish + start of a subscribe callback seems oddly slow.
I want to have multiple system components as objects in my workspace such that they respond to ROS topic updates or on timer events. The pc used for testing is not a monster but should not be garbage either.
Do you also think the shown time requirements are unneccesary large? this allows barely to publish a single topic at 200hz without doing anything else. Normally I have multiple lower frequency topics (e.g.20hz) but the total consumed time becomes significant.
Do you know any practices to make the system operate quicker?
What do you think of the OOP style of making control system components in general?
classdef subpubspeedMonitor < handle
% Use: call in matlab console, after initializing ros:
%
% SPM1 = subpubspeedMonitor()
%
% This will create an object which starts a set repetitive task upon creation
% and finally destructs itself after posting results in console.
properties
node
subscriber
publisher
timestart
messagetotal
end
methods
function obj = subpubspeedMonitor()
obj.node = ros.Node('subspeedmonitor1');
obj.subscriber = ros.Subscriber(obj.node,'topic1','sensor_msgs/NavSatFix',{#obj.rosSubCallback});
obj.publisher = ros.Publisher(obj.node,'topic1','sensor_msgs/NavSatFix');
obj.timestart = tic;
obj.messagetotal = 0;
msg1 = rosmessage(obj.publisher);
% Choose to evaluate subscriber + publisher loop or just counting
if 1
send(obj.publisher,msg1);
else
countAndDisplay(obj)
end
end
%% Test method one: repetitive publishing and subscribing
function rosSubCallback(obj,~,msg_) % ~3.7 ms per loop for a simple publish+subscribe action
% Latency to self is 0.05ms on average, according to "pinging" in terminal
obj.messagetotal = obj.messagetotal+1;
if obj.messagetotal <10000
%msg1 = rosmessage(obj.publisher); % this line adds 4.3000ms per loop
msg_.Longitude = 51; % this line adds 0.25000 ms per loop
send(obj.publisher,msg_)
else
% Display some results
timepassed = toc(obj.timestart);
time_per_pubsub = timepassed/obj.messagetotal
delete(obj);
end
end
%% Test method two: simply counting
function countAndDisplay(obj) % this costs 2.1318e-03 ms(!) per loop
obj.messagetotal = obj.messagetotal+1;
if obj.messagetotal <10000
%msg1 = rosmessage(obj.publisher); %adds 3.6ms per loop
%i = 1% adds 5.7532e-03 ms per loop
%msg1 = rosmessage("std_msgs/Bool"); %adds 1.5ms per loop
countAndDisplay(obj);
else
% Display some results
timepassed = toc(obj.timestart);
time_per_count_FCN = timepassed/obj.messagetotal
delete(obj);
end
end
%% Deconstructor
function delete(obj)
delete(obj.subscriber)
delete(obj.publisher)
delete(obj.node)
end
end
end
Assertion of condition is well known way to design application in strategic way. You could be completely sure that your code will work correctly a day after release, but also when other dev in your team will change this code.
There are 2 common ways to put assertion in Lua code:
assert(1 > 0, "Assert that math works")
if 1 <= 0 then
error("Assert that math doesn't work")
end
I would expect this thing similar from performance point of view.
Consider it only matter of style. But it happens to be not true.
assert works longer on my machine:
function with_assert()
for i=1,100000 do
assert(1 < 0, 'Assert')
end
end
function with_error()
for i=1,100000 do
if 1 > 0 then
error('Error')
end
end
end
local t = os.clock()
pcall(with_assert)
print(os.clock() - t)
t = os.clock()
pcall(with_error)
print(os.clock() - t)
>> 3.1999999999999e-05
>> 1.5e-05
Why does it happen?
Look at the source code of assert and error: assert does some work and then calls error.
But the loop in your code serves no purpose: throwing an error during the first iteration means that the rest of the iterations don't run. Perhaps you meant to put the loop around the pcalls as below. Then you'll find hardly any difference between the times.
function with_assert()
assert(1 < 0, 'Assert')
end
function with_error()
if 1 > 0 then
error('Error')
end
end
function test(f)
local t = os.clock()
for i=1,100000 do
pcall(f)
end
print(os.clock() - t)
end
test(with_assert)
test(with_error)
I have made that loop my self and Iam trying to make it faster, better... but sometimes after it repeat searching for existing... it press random ( i think cuz its not similar to any img iam using in sikuli ) place on the screen. Maybe you will know why.
Part of this loop below
while surowiec_1:
if exists("1451060448708.png", 1) or exists("1451061746632.png", 1):
foo = [w_lewo, w_prawo, w_dol, w_gore]
randomListElement = foo[random.randint(0,len(foo)-1)]
click(randomListElement)
wait(3)
else:
if exists("1450930340868.png", 1 ):
click(hemp)
wait(1)
hemp = exists("1450930340868.png", 1)
elif exists("1451086210167.png", 1):
click(tree)
wait(1)
tree = exists("1451086210167.png", 1)
elif exists("1451022614047.png", 1 ):
hover("1451022614047.png")
click(flower)
flower = exists("1451022614047.png", 1)
elif exists("1451021823366.png", 1 ):
click(fish)
fish = exists("1451021823366.png")
elif exists("1451022083851.png", 1 ):
click(bigfish)
bigfish = exists("1451022083851.png", 1)
else:
foo = [w_lewo, w_prawo, w_dol, w_gore]
randomListElement = foo[random.randint(0,len(foo)-1)]
click(randomListElement)
wait(3)
I wonder if this is just program problem with img recognitions or I have made a mistake.
You call twice the exist method indending to get the same match (the first one in your if statement, the second time to assign it to the value. You ask sikuli to evaluate the image twice, and it can have different results.
From the method's documentation
the best match can be accessed using Region.getLastMatch() afterwards.
if volt.isalpha() or res.isalpha() or amp.isalpha():
What did I do wrong here? I get an INVALID SYNTAX, I am using this for a calculator program I am making. It calculates voltage, resistance, and amperage. But thats the easy part, I am just trying to make it fool proof. I have 3 variables in the code (volt, amp, res) that are inputted by the user. I just wanna make sure that they don't type in anything stupid. Like letters for e.g. ...
try:
float(volt) >= 0 and float(res) >= 0 and float(amp) >= 0
print("")
print("You put a value for everything. You don't need the calculator.")
allowed = 0
if volt.isalpha() or res.isalpha() or amp.isalpha():
print("You typed in characters for one of the values, this calculator doesn't use letters.")
allowed = 0
def find_voltage(a,b): # V = I * R
voltage = a * b
return(voltage)`
You don't have an except block after try - it is required. Do something like:
try:
float(volt) >= 0 and float(res) >= 0 and float(amp) >= 0
print("")
print("You put a value for everything. You don't need the calculator.")
allowed = 0
except ValueError:
print("Oops, you messed up.")
Additionally, the line
float(volt) >= 0 and float(res) >= 0 and float(amp) >= 0
doesn't do anything. You'll need to assign it to a variable, then check the results of the variable - if True, do one thing, if False, do something else.
I have recently learned how to work with basic files in Fortran
and I assumed it was as simple as:
open(unit=10,file="data.dat")
read(10,*) some_variable, somevar2
close(10)
So I can't understand why this function I wrote is not working.
It compiles fine but when I run it it prints:
fortran runtime error:end of file
Code:
Function Load_Names()
character(len=30) :: Staff_Name(65)
integer :: i = 1
open(unit=10, file="Staff_Names.txt")
do while(i < 65)
read(10,*) Staff_Name(i)
print*, Staff_Name(i)
i = i + 1
end do
close(10)
end Function Load_Names
I am using Fortran 2008 with gfortran.
A common reason for the error you report is that the program doesn't find the file it is trying to open. Sometimes your assumptions about the directory in which the program looks for files at run-time will be wrong.
Try:
using the err= option in the open statement to write code to deal gracefully with a missing file; without this the program crashes, as you have observed;
or
using the inquire statement to figure out whether the file exists where your program is looking for it.
You can check when a file has ended. It is done with the option IOSTAT for read statement.
Try:
Function Load_Names()
character(len=30) :: Staff_Name(65)
integer :: i = 1
integer :: iostat
open(unit=10, file="Staff_Names.txt")
do while(i < 65)
read(10,*, IOSTAT=iostat) Staff_Name(i)
if( iostat < 0 )then
write(6,'(A)') 'Warning: File containts less than 65 entries'
exit
else if( iostat > 0 )then
write(6,'(A)') 'Error: error reading file'
stop
end if
print*, Staff_Name(i)
i = i + 1
end do
close(10)
end Function Load_Names
Using Fortran 2003 standard, one can do the following to check if the end of file is reached:
use :: iso_fortran_env
character(len=1024) :: line
integer :: u1,stat
open (newunit=u1,action='read',file='input.dat',status='old')
ef: do
read(u1,'A',iostat=stat) line
if (stat == iostat_end) exit ef ! end of file
...
end do ef
close(u1)
Thanks for all your help i did fix the code:
Function Load_Names(Staff_Name(65))!Loads Staff Names
character(len=30) :: Staff_Name(65)
integer :: i = 1
open(unit=10, file="Staff_Names.txt", status='old', action='read')!opens file for reading
do while(i < 66)!Sets Set_Name() equal to the file one string at a time
read(10,*,end=100) Staff_Name(i)
i = i + 1
end do
100 close(10)!closes file
return!returns Value
end Function Load_Names
I needed to change read(10,*) to read(10,*,END=100)
so it knew what to do when it came to the end the file
as it was in a loop I assume.
Then your problem was that your file was a row vector, and it was likely
giving you this error immediately after reading the first element, as #M.S.B. was suggesting.
If you have a file with a NxM matrix and you read it in this way (F77):
DO i=1,N
DO j=1,M
READ(UNIT,*) Matrix(i,j)
ENDDO
ENDDO
it will load the first column of your file in the first row of your matrix and will give you an error as soon as it reaches the end of the file's first column, because the loop enforces it to read further lines and there are no more lines (if N<M when j=N+1 for example). To read the different columns you should use an implicit loop, which is why your solution worked:
DO i=1,N
READ(UNIT,*) (Matrix(i,j), j=1,M)
ENDDO
I am using GNU Fortran 5.4.0 on the Ubuntu system 16.04. Please check your file if it is the right one you are looking for, because sometimes files of the same name are confusing, and maybe one of them is blank. As you may check the file path if it is in the same working directory.