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  • boost thread pool

    - by Dtag
    I need a threadpool for my application, and I'd like to rely on standard (C++11 or boost) stuff as much as possible. I realize there is an unofficial(!) boost thread pool class, which basically solves what I need, however I'd rather avoid it because it is not in the boost library itself -- why is it still not in the core library after so many years? In some posts on this page and elsewhere, people suggested using boost::asio to achieve a threadpool like behavior. At first sight, that looked like what I wanted to do, however I found out that all implementations I have seen have no means to join on the currently active tasks, which makes it useless for my application. To perform a join, they send stop signal to all the threads and subsequently join them. However, that completely nullifies the advantage of threadpools in my use case, because that makes new tasks require the creation of a new thread. What I want to do is: ThreadPool pool(4); for (...) { for (int i=0;i<something;i++) pool.pushTask(...); pool.join(); // do something with the results } Can anyone suggest a solution (except for using the existing unofficial thread pool on sourceforge)? Is there anything in C++11 or core boost that can help me here? Thanks a lot

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  • Sorting a meta-list by first element of children lists in Python

    - by thismachinechills
    I have a list, root, of lists, root[child0], root[child1], etc. I want to sort the children of the root list by the first value in the child list, root[child0][0], which is an int. Example: import random children = 10 root = [[random.randint(0, children), "some value"] for child in children] I want to sort root from greatest to least by the first element of each of it's children. I've taken a look at some previous entries that used sorted() and a lamda function I'm entirely unfamiliar with, so I'm unsure of how to apply that to my problem. Appreciate any direction that can by given Thanks

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  • c++/boost: use tuple ctors when subclassing

    - by bbb
    Hi there, is there some way to use a boost tuple's ctors as an addition to the subclass methods (and ctors) like here? // typedef boost::tuple<int, SomeId, SomeStatus> Conn; // Conn(1); // works and initializes using default ctors of Some* struct Conn : boost::tuple<int, AsynchId, AccDevRetStatus> {}; Conn(1); // "no matching function call" (but i want it so much) T.H.X.

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  • Python 3: Recursivley find if number is even

    - by pythonhack
    I am writing a program that must find if a number is even or not. It needs to follow this template. I can get it to find if a number is even or not recursively (call function and subtract 2, base case zero), but I am having a hard time following this template, based on how the isEven function is called in the main function. Any help would be greatly appreciated. Write a recursive function called isEven that finds whether a number is even or not: def isEven() #recursivley determine whether number is even or not def main(): number=int(input(“Enter a number : “)) if (isEven(number)): print(“Number is even”) else: print(“Number is not even”) main() Thank you! Appreciate it.

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  • Another boost error

    - by user1676605
    On this code I get the enourmous error static void ParseTheCommandLine(int argc, char *argv[]) { int count; int seqNumber; namespace po = boost::program_options; std::string appName = boost::filesystem::basename(argv[0]); po::options_description desc("Generic options"); desc.add_options() ("version,v", "print version string") ("help", "produce help message") ("sequence-number", po::value<int>(&seqNumber)->default_value(0), "sequence number") ("pem-file", po::value< vector<string> >(), "pem file") ; po::positional_options_description p; p.add("pem-file", -1); po::variables_map vm; po::store(po::command_line_parser(argc, argv). options(desc).positional(p).run(), vm); po::notify(vm); if (vm.count("pem file")) { cout << "Pem files are: " << vm["pem-file"].as< vector<string> >() << "\n"; } cout << "Sequence number is " << seqNumber << "\n"; exit(1); ../../../FIXMarketDataCommandLineParameters/FIXMarketDataCommandLineParameters.hpp|98|error: no match for ‘operator<<’ in ‘std::operator<< [with _Traits = std::char_traits](((std::basic_ostream &)(& std::cout)), ((const char*)"Pem files are: ")) << ((const boost::program_options::variable_value*)vm.boost::program_options::variables_map::operator[](((const std::string&)(& std::basic_string, std::allocator (((const char*)"pem-file"), ((const std::allocator&)((const std::allocator*)(& std::allocator()))))))))-boost::program_options::variable_value::as with T = std::vector, std::allocator , std::allocator, std::allocator ’|

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  • threading in Python taking up too much CPU

    - by KevinShaffer
    I wrote a chat program and have a GUI running using Tkinter, and to go and check when new messages have arrived, I create a new thread so Tkinter keeps doing its thing without locking up while the new thread goes and grabs what I need and updates the Tkinter window. This however becomes a huge CPU hog, and my guess is that it has to do somehow with the fact that the Thread is started and never really released when the function is done. Here's the relevant code (it's ugly and not optimized at the moment, but it gets the job done, and itself does not use too much processing power, as when I run it not threaded, it doesn't take up much CPU but it locks up Tkinter) Note: This is inside of a class, hence the extra tab. def interim(self): threading.Thread(target=self.readLog).start() self.after(5000,self.interim) def readLog(self): print 'reading' try: length = len(str(self.readNumber)) f = open('chatlog'+str(myport),'r') temp = f.readline().replace('\n','') while (temp[:length] != str(self.readNumber)) or temp[0] == '<': temp = f.readline().replace('\n','') while temp: if temp[0] != '<': self.updateChat(temp[length:]) self.readNumber +=1 else: self.updateChat(temp) temp = f.readline().replace('\n','') f.close() Is there a way to better manage the threading so I don't consume 100% of the CPU very quickly?

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  • boost::asio::io_service throws exception

    - by Ace
    Okay, I seriously cannot figure this out. I have a DLL project in MSVC that is attempting to use Asio (from Boost 1.45.0), but whenever I create my io_service, an exception is thrown. Here is what I am doing for testing purposes: void run() { boost::this_thread::sleep(boost::posix_time::seconds(5)); try { boost::asio::io_service io_service; } catch (std::exception & e) { MessageBox(NULL, e.what(), "Exception", MB_OK); } } BOOL WINAPI DllMain(HINSTANCE hinstDLL, DWORD fdwReason, LPVOID lpvReserved) { if (fdwReason == DLL_PROCESS_ATTACH) { boost::thread thread(run); } return TRUE; } This is what the message box shows: winsock: WSAStartup cannot function at this time because the underlying system it uses to provide network services is currently unavailable Here is what MSDN says about it (error code 10091, WSASYSNOTREADY): Network subsystem is unavailable. This error is returned by WSAStartup if the Windows Sockets implementation cannot function at because the underlying system it uses to provide network services is currently unavailable. Users should check: That the appropriate Windows Sockets DLL file is in the current path. That they are not trying to use more than one Windows Sockets implementation simultaneously. If there is more than one Winsock DLL on your system, be sure the first one in the path is appropriate for the network subsystem currently loaded. The Windows Sockets implementation documentation to be sure all necessary components are currently installed and configured correctly. Yet none of this seems to apply to me (or so I think). Here is my command line: /O2 /GL /D "_WIN32_WINNT=0x0501" /D "_WINDLL" /FD /EHsc /MD /Gy /Fo"Release\" /Fd"Release\vc90.pdb" /W3 /WX /nologo /c /TP /errorReport:prompt If anyone knows what might be wrong, please help me out! Thanks.

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  • Python 3 order of testing undetermined

    - by user578598
    string='a' p=0 while (p <len(string)) & (string[p]!='c') : p +=1 print ('the end but the process already died ') while (p <1) & (string[p]!='c') : IndexError: string index out of range I want to test a condition up to the end of a string (example string length=1) why are both parts of the and executed is the condition is already false! as long as p < len(string). the second part does not even need executing. if it does a lot of performance can be lost

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  • Compilation failing - no #include - boost

    - by jwoolard
    Hi, I'm trying to compile a third-party library, but g++ is complaining about the following line: typedef boost::shared_ptr<MessageConsumer> MessageConsumerPtr; The strange thing is, there is no #include directive in the file - and it is clearly supposed to be this way; there are about 60 files with the same (or very similar) issues. Clearly if there was an #include directive referencing the relevant boost header this would compile cleanly. My question is: how can I get g++ to somehow automagically find the relevant symbol (in all instances of this issue, it is a namespace that can't be found - usually std:: or boost::) by either automatically processing the relevant header (or some other mechanism). Thanks. Edit My current g++ call looks like: g++ -fPIC -O3 -DUSING_PCH -D_REENTRANT -I/usr/include/boost -I./ -c MessageInterpreter.cpp -o MessageInterpreter.o

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  • compiling Boost linked libraries (Ubuntu)

    - by Adam Greenhall
    I installed Boost via sudo apt-get install libboost-all-dev on the most recent version of Ubuntu. Now I want to compile a project that uses the Boost.Serialization library, which needs to be linked. I've tried many variants of the following, without success: gcc -I /usr/lib code.cpp -o compiled /usr/lib/libboost_serialization.a and gcc -I /usr/lib code.cpp -o compiled -l libboost_serialization The error message is: error: ‘split_member’ is not a member of ‘boost::serialization ` What am I missing?

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  • Project Euler 10: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 10.  As always, any feedback is welcome. # Euler 10 # http://projecteuler.net/index.php?section=problems&id=10 # The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. # Find the sum of all the primes below two million. import time start = time.time() def primes_to_max(max): primes, number = [2], 3 while number < max: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes primes = primes_to_max(2000000) print sum(primes) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • What's wrong with relative imports in Python?

    - by Oddthinking
    I recently upgraded versions of pylint, a popular Python style-checker. It has gone ballistic throughout my code, pointing out places where I import modules in the same package, without specifying the full package path. The new error message is W0403. W0403: Relative import %r, should be %r Used when an import relative to the package directory is detected. Example For example, if my packages are structured like this: /cake /__init__.py /icing.py /sponge.py /drink and in the sponge package I write: import icing instead of import cake.icing I will get this error. While I understand that not all Pylint messages are of equal importance, and I am not afraid to dismiss them, I don't understand why such a practice is considered a poor idea. I was hoping someone could explain the pitfalls, so I could improve my coding style rather than (as I currently plan to do) turning off this apparently spurious warning.

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  • Project Euler 15: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 15.  As always, any feedback is welcome. # Euler 15 # http://projecteuler.net/index.php?section=problems&id=15 # Starting in the top left corner of a 2x2 grid, there # are 6 routes (without backtracking) to the bottom right # corner. How many routes are their in a 20x20 grid? import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) rows, cols = 20, 20 print factorial(rows+cols) / (factorial(rows) * factorial(cols)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 9: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 9.  As always, any feedback is welcome. # Euler 9 # http://projecteuler.net/index.php?section=problems&id=9 # A Pythagorean triplet is a set of three natural numbers, # a b c, for which, # a2 + b2 = c2 # For example, 32 + 42 = 9 + 16 = 25 = 52. # There exists exactly one Pythagorean triplet for which # a + b + c = 1000. Find the product abc. import time start = time.time() product = 0 def pythagorean_triplet(): for a in range(1,501): for b in xrange(a+1,501): c = 1000 - a - b if (a*a + b*b == c*c): return a*b*c print pythagorean_triplet() print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 5: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 5.  As always, any feedback is welcome. # Euler 5 # http://projecteuler.net/index.php?section=problems&id=5 # 2520 is the smallest number that can be divided by each # of the numbers from 1 to 10 without any remainder. # What is the smallest positive number that is evenly # divisible by all of the numbers from 1 to 20? import time start = time.time() def gcd(a, b): while b: a, b = b, a % b return a def lcm(a, b): return a * b // gcd(a, b) print reduce(lcm, range(1, 20)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Compound assignment operators in Python's Numpy library

    - by Leonard
    The "vectorizing" of fancy indexing by Python's numpy library sometimes gives unexpected results. For example: import numpy a = numpy.zeros((1000,4), dtype='uint32') b = numpy.zeros((1000,4), dtype='uint32') i = numpy.random.random_integers(0,999,1000) j = numpy.random.random_integers(0,3,1000) a[i,j] += 1 for k in xrange(1000): b[i[k],j[k]] += 1 Gives different results in the arrays 'a' and 'b' (i.e. the appearance of tuple (i,j) appears as 1 in 'a' regardless of repeats, whereas repeats are counted in 'b'). This is easily verified as follows: numpy.sum(a) 883 numpy.sum(b) 1000 It is also notable that the fancy indexing version is almost two orders of magnitude faster than the for loop. My question is: "Is there an efficient way for numpy to compute the repeat counts as implemented using the for loop in the provided example?"

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  • Project Euler 8: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 8.  As always, any feedback is welcome. # Euler 8 # http://projecteuler.net/index.php?section=problems&id=8 # Find the greatest product of five consecutive digits # in the following 1000-digit number import time start = time.time() number = '\ 73167176531330624919225119674426574742355349194934\ 96983520312774506326239578318016984801869478851843\ 85861560789112949495459501737958331952853208805511\ 12540698747158523863050715693290963295227443043557\ 66896648950445244523161731856403098711121722383113\ 62229893423380308135336276614282806444486645238749\ 30358907296290491560440772390713810515859307960866\ 70172427121883998797908792274921901699720888093776\ 65727333001053367881220235421809751254540594752243\ 52584907711670556013604839586446706324415722155397\ 53697817977846174064955149290862569321978468622482\ 83972241375657056057490261407972968652414535100474\ 82166370484403199890008895243450658541227588666881\ 16427171479924442928230863465674813919123162824586\ 17866458359124566529476545682848912883142607690042\ 24219022671055626321111109370544217506941658960408\ 07198403850962455444362981230987879927244284909188\ 84580156166097919133875499200524063689912560717606\ 05886116467109405077541002256983155200055935729725\ 71636269561882670428252483600823257530420752963450' max = 0 for i in xrange(0, len(number) - 5): nums = [int(x) for x in number[i:i+5]] val = reduce(lambda agg, x: agg*x, nums) if val > max: max = val print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Python or C server hosting for indie development

    - by Richard Fabian
    I've written a lot of the game, but it's singleplayer. Now we want to join up and play together. I want to host it like an MMO, but haven't got any personal ability to host (no static IPs or direct access to a reasonable router that will allow me to port forward) so I wondered if there were any free / very cheap hosting solutions for people developing games that need to develop their MMO side. In my case it's a world server for a 2D game where the world map can be changed by the players. So, GAE sounds expensive, as there would be quite a few updates per second (I heard they bill for data updates but not for download, but can't find refernce to billing anywhere on the FAQs) I'd prefer to be able to write the server in python as that's what the game is written in (with pygame), but C is fine, and maybe even better as it might prompt me to write some more performant world generator code ;)

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  • Project Euler 2: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 2.  As always, any feedback is welcome. # Euler 2 # http://projecteuler.net/index.php?section=problems&id=2 # Find the sum of all the even-valued terms in the # Fibonacci sequence which do not exceed four million. # Each new term in the Fibonacci sequence is generated # by adding the previous two terms. By starting with 1 # and 2, the first 10 terms will be: # 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... # Find the sum of all the even-valued terms in the # sequence which do not exceed four million. import time start = time.time() total = 0 previous = 0 i = 1 while i <= 4000000: if i % 2 == 0: total +=i # variable swapping removes the need for a temp variable i, previous = previous, previous + i print total print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • PHP developer wanting to learn python

    - by dclowd9901
    I'm pretty familiar at this point with PHP (Javascript, too), up to the point of OOP in PHP, and am looking to branch out my knowledge. I'm looking at Python next, but a lot of it is a bit alien to me as a PHP developer. I'm less concerned about learning the language itself. I'm positive there's plenty of good resources, documentation and libraries to help me get the code down. I'm less sure about the technical aspects of how to set up a dev environment, unit testing and other more mundane details that are very important, aid in rapid development, but aren't as widely covered. Are there any good resources out there for this?

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  • Project Euler 16: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 16.  As always, any feedback is welcome. # Euler 16 # http://projecteuler.net/index.php?section=problems&id=16 # 2^15 = 32768 and the sum of its digits is # 3 + 2 + 7 + 6 + 8 = 26. # What is the sum of the digits of the number 2^1000? import time start = time.time() print sum([int(i) for i in str(2**1000)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 13: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 13.  As always, any feedback is welcome. # Euler 13 # http://projecteuler.net/index.php?section=problems&id=13 # Work out the first ten digits of the sum of the # following one-hundred 50-digit numbers. import time start = time.time() number_string = '\ 37107287533902102798797998220837590246510135740250\ 46376937677490009712648124896970078050417018260538\ 74324986199524741059474233309513058123726617309629\ 91942213363574161572522430563301811072406154908250\ 23067588207539346171171980310421047513778063246676\ 89261670696623633820136378418383684178734361726757\ 28112879812849979408065481931592621691275889832738\ 44274228917432520321923589422876796487670272189318\ 47451445736001306439091167216856844588711603153276\ 70386486105843025439939619828917593665686757934951\ 62176457141856560629502157223196586755079324193331\ 64906352462741904929101432445813822663347944758178\ 92575867718337217661963751590579239728245598838407\ 58203565325359399008402633568948830189458628227828\ 80181199384826282014278194139940567587151170094390\ 35398664372827112653829987240784473053190104293586\ 86515506006295864861532075273371959191420517255829\ 71693888707715466499115593487603532921714970056938\ 54370070576826684624621495650076471787294438377604\ 53282654108756828443191190634694037855217779295145\ 36123272525000296071075082563815656710885258350721\ 45876576172410976447339110607218265236877223636045\ 17423706905851860660448207621209813287860733969412\ 81142660418086830619328460811191061556940512689692\ 51934325451728388641918047049293215058642563049483\ 62467221648435076201727918039944693004732956340691\ 15732444386908125794514089057706229429197107928209\ 55037687525678773091862540744969844508330393682126\ 18336384825330154686196124348767681297534375946515\ 80386287592878490201521685554828717201219257766954\ 78182833757993103614740356856449095527097864797581\ 16726320100436897842553539920931837441497806860984\ 48403098129077791799088218795327364475675590848030\ 87086987551392711854517078544161852424320693150332\ 59959406895756536782107074926966537676326235447210\ 69793950679652694742597709739166693763042633987085\ 41052684708299085211399427365734116182760315001271\ 65378607361501080857009149939512557028198746004375\ 35829035317434717326932123578154982629742552737307\ 94953759765105305946966067683156574377167401875275\ 88902802571733229619176668713819931811048770190271\ 25267680276078003013678680992525463401061632866526\ 36270218540497705585629946580636237993140746255962\ 24074486908231174977792365466257246923322810917141\ 91430288197103288597806669760892938638285025333403\ 34413065578016127815921815005561868836468420090470\ 23053081172816430487623791969842487255036638784583\ 11487696932154902810424020138335124462181441773470\ 63783299490636259666498587618221225225512486764533\ 67720186971698544312419572409913959008952310058822\ 95548255300263520781532296796249481641953868218774\ 76085327132285723110424803456124867697064507995236\ 37774242535411291684276865538926205024910326572967\ 23701913275725675285653248258265463092207058596522\ 29798860272258331913126375147341994889534765745501\ 18495701454879288984856827726077713721403798879715\ 38298203783031473527721580348144513491373226651381\ 34829543829199918180278916522431027392251122869539\ 40957953066405232632538044100059654939159879593635\ 29746152185502371307642255121183693803580388584903\ 41698116222072977186158236678424689157993532961922\ 62467957194401269043877107275048102390895523597457\ 23189706772547915061505504953922979530901129967519\ 86188088225875314529584099251203829009407770775672\ 11306739708304724483816533873502340845647058077308\ 82959174767140363198008187129011875491310547126581\ 97623331044818386269515456334926366572897563400500\ 42846280183517070527831839425882145521227251250327\ 55121603546981200581762165212827652751691296897789\ 32238195734329339946437501907836945765883352399886\ 75506164965184775180738168837861091527357929701337\ 62177842752192623401942399639168044983993173312731\ 32924185707147349566916674687634660915035914677504\ 99518671430235219628894890102423325116913619626622\ 73267460800591547471830798392868535206946944540724\ 76841822524674417161514036427982273348055556214818\ 97142617910342598647204516893989422179826088076852\ 87783646182799346313767754307809363333018982642090\ 10848802521674670883215120185883543223812876952786\ 71329612474782464538636993009049310363619763878039\ 62184073572399794223406235393808339651327408011116\ 66627891981488087797941876876144230030984490851411\ 60661826293682836764744779239180335110989069790714\ 85786944089552990653640447425576083659976645795096\ 66024396409905389607120198219976047599490197230297\ 64913982680032973156037120041377903785566085089252\ 16730939319872750275468906903707539413042652315011\ 94809377245048795150954100921645863754710598436791\ 78639167021187492431995700641917969777599028300699\ 15368713711936614952811305876380278410754449733078\ 40789923115535562561142322423255033685442488917353\ 44889911501440648020369068063960672322193204149535\ 41503128880339536053299340368006977710650566631954\ 81234880673210146739058568557934581403627822703280\ 82616570773948327592232845941706525094512325230608\ 22918802058777319719839450180888072429661980811197\ 77158542502016545090413245809786882778948721859617\ 72107838435069186155435662884062257473692284509516\ 20849603980134001723930671666823555245252804609722\ 53503534226472524250874054075591789781264330331690' total = 0 for i in xrange(0, 100 * 50 - 1, 50): total += int(number_string[i:i+49]) print str(total)[:10] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Finding the html tag value with Python [on hold]

    - by MrWho
    Consider a html page, which contains a line like below: file: 'http://google.com/video.mp4' I want to search for google.com/video.mp4 in that file and save it in a variable.I want to code it with python. Shortly, I want to elicit a link from a html page, so I need to get the link by using regular expressions or the other techniques in which I'm asking about. PS: What should I exactly try to clarify?it's really annoying that the administrators don't even say what is exactly unclear about the question, they've just learned to close or on hold the topics!

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  • Project Euler 7: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 7.  As always, any feedback is welcome. # Euler 7 # http://projecteuler.net/index.php?section=problems&id=7 # By listing the first six prime numbers: 2, 3, 5, 7, # 11, and 13, we can see that the 6th prime is 13. What # is the 10001st prime number? import time start = time.time() def nthPrime(nth): primes = [2] number = 3 while len(primes) < nth: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes[nth - 1] print nthPrime(10001) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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