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  • iphone problem receiving UDP packets

    - by SooDesuNe
    I'm using sendto() and recvfrom() to send some simple packets via UDP over WiFI. I've tried using two phones, and a simulator, the results I'm getting are: Packets sent from phones - recieved by simulator Packets sent from simulator - simulator recvfrom remains blocking. Packets sent from phones - other phone recvfrom remains blocking. I'm not sure how to start debugging this one, since the simulator/mac is able to receive the the packets, but the phones don't appear to be getting the message. A slight aside, do I need to keep my packets below the MTU for my network? Or is fragmentation handled by the OS or some other lower level software? UPDATE: I forgot to include the packet size and structure. I'm transmitting: typedef struct PacketForTransmission { int32_t packetTypeIdentifier; char data[64]; // size to fit my biggest struct } PacketForTransmission; of which the char data[64] is: typedef struct PacketHeader{ uint32_t identifier; uint32_t datatype; } PacketHeader; typedef struct BasePacket{ PacketHeader header; int32_t cardValue; char sendingDeviceID[41]; //dont forget to save room for the NULL terminator! } BasePacket; typedef struct PositionPacket{ BasePacket basePacket; int32_t x; int32_t y; } PositionPacket; sending packet is like: PositionPacket packet; bzero(&packet, sizeof(packet)); //fill packet with it's associated data PacketForTransmission transmissionPacket; transmissionPacket.packetTypeIdentifier = kPositionPacketType; memcpy(&transmissionPacket.data, (void*)&packet, sizeof(packet)); //put the PositionPacket into data[64] size_t sendResult = sendto(_socket, &transmissionPacket, sizeof(transmissionPacket), 0, [address bytes], [address length]); NSLog(@"packet sent of size: %i", sendResult); and recieving packets is like: while(1){ char dataBuffer[8192]; struct sockaddr addr; socklen_t socklen = sizeof(addr); ssize_t len = recvfrom(_socket, dataBuffer, sizeof(dataBuffer), 0, &addr, &socklen); //continues blocking here NSLog(@"packet recieved of length: %i", len); //do some more stuff }

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  • Loop through children and display each, as3

    - by VideoDnd
    How do I loop through all of my children, and display each? I would like to know the best way to do this. my children and containerfive children, one plays every sec, 1,2,3, etc. var square1:Square1 = new Square1; var square2:Square2 = new Square2; var square3:Square3 = new Square3; var square4:Square4 = new Square4; var square5:Square5 = new Square5; var container:Sprite = new Sprite; addChild(container); container.addChild(square1) container.addChild(square2) container.addChild(square3) container.addChild(square4) container.addChild(square5) my timer var timly:Timer = new Timer(1000, 5); timly.start(); timly.addEventListener(TimerEvent.TIMER, onLoop); Note: Tried for loop, numChildren -1, and visibility ERROR 'access of undefined property' //Thomas's idea var timly:Timer = new Timer(1000, 10); timly.start(); timly.addEventListener(TimerEvent.TIMER, onLoop, false, 0, true); // var square1:Square1 = new Square1; square1.visible = false container.addChild(square2) var square2:Square2 = new Square2; square2.visible = false container.addChild(square3) var square3:Square3 = new Square3; square3.visible = false container.addChild(square3) var square4:Square4 = new Square4; square4.visible = false container.addChild(square4) var square5:Square5 = new Square5; square5.visible = false container.addChild(square5) var container:Sprite = new Sprite; this.addChild(container); var curCount:Number = 100; // function collectChildren(container:DisplayObjectContainer):Array { var len:int = container.numChildren; var mySquaresArray:Array = []; for (var i:int = 0; i < len; i++) { mySquaresArray.push(container.getChildAt(i).name); } return mySquaresArray; } // function onLoop( e:Event ) { curCount = e.target.currentCount; if( curCount > 1 ) { var previous_square = curCount -2; mySquaresArray[previous_square].visible = false; } var current_square = curCount - 1; mySquaresArray[current_square].visible = true; }

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  • Unknown syntax error.

    - by matt1024
    Why do I get a syntax error running this code? If I remove the highlighted section (return cards[i]) I get the error highlighting the function call instead. Please help :) def dealcards(): for i in range(len(cards)): cards[i] = '' for j in range(8): cards[i] = cards[i].append(random.randint(0,9) return cards[i] print (dealcards())

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  • SQL column length query

    - by Mike
    i've been using the following query: select LEN(columnname) as columnmame from dbo.amu_datastaging This works, but is there a way to only return the greatest value instead of all the values? So if i return 1million records and the longest length is 400, the query would just return the value of 400?

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  • MS Access ADODB.recordset character limit is 2036!? Can this be increased?

    - by souper-dragon
    In the following AccessVBA code, I am trying to write a record to a memo field called "Recipient_Display": oRec1.Fields("RECIPIENT_DISPLAY") = Left(sRecipientDisplayNames, Len(sRecipientDisplayNames) - 2) When the string contains 2036 characters, the write completes. Above this number I get the following error: Run-time error'-2147217887(80040e21)': Could not update; currently locked by another session on this machine. What is the significance of this number 2036 and is there a property I can adjust that will allow the above update to take place?

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  • Java: Is there a way to obtain the bytecode for a class at runtime?

    - by Adam Paynter
    In Java, is there a way (at runtime) to obtain the bytecode which defined a particular class? Put another way, is there a way to obtain the byte[] array passed to ClassLoader.defineClass(String name, byte[] b, int off, int len) when a particular class was loaded? I see that this method is declared final, so creating a custom ClassLoader to intercept class definitions seems out of the question. In the past, I have used the class's ClassLoader to obtain the bytecode via the getResourceAsStream(String) method, but I would prefer a more canonical solution.

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  • compare two following values in numpy array

    - by Billy Mitchell
    What is the best way to touch two following values in an numpy array? example: npdata = np.array([13,15,20,25]) for i in range( len(npdata) ): print npdata[i] - npdata[i+1] this looks really messed up and additionally needs exception code for the last iteration of the loop. any ideas? Thanks!

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  • itertools.product eliminating repeated reversed tuples

    - by genclik27
    I asked a question yesterday and thanks to Tim Peters, it is solved. The question is here; itertools.product eliminating repeated elements The new question is further version of this. This time I will generate tuples inside of tuples. Here is an example; lis = [[(1,2), (3,4)], [(5,2), (1,2)], [(2,1), (1,2)]] When I use it in itertools.product function this is what I get, ((1, 2), (5, 2), (2, 1)) ((1, 2), (5, 2), (1, 2)) ((1, 2), (1, 2), (2, 1)) ((1, 2), (1, 2), (1, 2)) ((3, 4), (5, 2), (2, 1)) ((3, 4), (5, 2), (1, 2)) ((3, 4), (1, 2), (2, 1)) ((3, 4), (1, 2), (1, 2)) I want to change it in a way that if a sequence has (a,b) inside of it, then it can not have (b,a). In this example if you look at this sequence ((3, 4), (1, 2), (2, 1)) it has (1,2) and (2,1) inside of it. So, this sequence ((3, 4), (1, 2), (2, 1)) should not be considered in the results. As I said, I asked similar question before, in that case it was not considering duplicate elements. I try to adapt it to my problem. Here is modified code. Changed parts in old version are taken in comments. def reverse_seq(seq): s = [] for i in range(len(seq)): s.append(seq[-i-1]) return tuple(s) def uprod(*seqs): def inner(i): if i == n: yield tuple(result) return for elt in sets[i] - reverse: #seen.add(elt) rvrs = reverse_seq(elt) reverse.add(rvrs) result[i] = elt for t in inner(i+1): yield t #seen.remove(elt) reverse.remove(rvrs) sets = [set(seq) for seq in seqs] n = len(sets) #seen = set() reverse = set() result = [None] * n for t in inner(0): yield t In my opinion this code should work but I am getting error for the input lis = [[(1,2), (3,4)], [(5,2), (1,2)], [(2,1), (1,2)]]. I could not understand where I am wrong. for i in uprod(*lis): print i Output is, ((1, 2), (1, 2), (1, 2)) Traceback (most recent call last): File "D:\Users\SUUSER\workspace tree\sequence_covering _array\denemeler_buraya.py", line 39, in <module> for i in uprod(*lis): File "D:\Users\SUUSER\workspace tree\sequence_covering _array\denemeler_buraya.py", line 32, in uprod for t in inner(0): File "D:\Users\SUUSER\workspace tree\sequence_covering _array\denemeler_buraya.py", line 22, in inner for t in inner(i+1): File "D:\Users\SUUSER\workspace tree\sequence_covering _array\denemeler_buraya.py", line 25, in inner reverse.remove(rvrs) KeyError: (2, 1) Thanks,

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  • Find last match with python regular expression

    - by SDD
    I wanto to match the last occurence of a simple pattern in a string, e.g. list = re.findall(r"\w+ AAAA \w+", "foo bar AAAA foo2 AAAA bar2) print "last match: ", list[len(list)-1] however, if the string is very long, a huge list of matches is generated. Is there a more direct way to match the second occurence of "AAAA" or should I use this workaround?

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  • Does this introduce security vulnerabilities?

    - by mcmt
    I don't think I'm missing anything. Then again I'm kind of a newbie. def GET(self, filename): name = urllib.unquote(filename) full = path.abspath(path.join(STATIC_PATH, filename)) #Make sure request is not tricksy and tries to get out of #the directory, e.g. filename = "../.ssh/id_rsa". GET OUTTA HERE assert full[:len(STATIC_PATH)] == STATIC_PATH, "bad path" return open(full).read()

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  • Python recursion with list returns None

    - by newman
    def foo(a): a.append(1) if len(a) > 10: print a return a else: foo(a) Why this recursive function returns None (see transcript below)? I can't quite understand what I am doing wrong. In [263]: x = [] In [264]: y = foo(x) [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] In [265]: print y None

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  • Why is a .net generic dictionary so big

    - by thefroatgt
    I am serializing a generic dictionary in VB.net and I am very surprised that it is about 1.3kb with a single item. Am I doing something wrong, or is there something else I should be doing? I have a large number of dictionaries and it is killing me to send them all across the wire. The code I use for serialization is Dim dictionary As New Dictionary(Of Integer, Integer) Dim stream As New MemoryStream Dim bformatter As New BinaryFormatter() dictionary.Add(1, 1) bformatter.Serialize(stream, dictionary) Dim len As Long = stream.Length

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  • AppEngine: how do cursors work?

    - by victor a.k.a. python for ever
    hello, i have the following code def get(self): date = datetime.date.today() loc_query = Location.all() last_cursor = memcache.get('location_cursor') if last_cursor: loc_query.with_cursor(last_cursor) loc_result = loc_query.fetch(1) for loc in loc_result: self.record(loc, date) taskqueue.add( url='/task/query/simplegeo', params={'date':date, 'locid':loc.key().id()} ) if len(loc_result): memcache.add('location_cursor', loc_query.cursor()) taskqueue.add(url='/task/count/', method='GET') else: memcache.add('location_cursor', None) i don't know what i'm doing wrong, but i am getting the same cursor which is not the effect i wanted. why isn't the cursor moving?

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  • A python code to convert a number from any base to the base of 10 giving errors . What is wrong with this code?

    - by mekasperasky
    import math def baseencode(number, base): ##Converting a number of any base to base10 if number == 0: return '0' for i in range(0,len(number)): if number[i]!= [A-Z]: num = num + number[i]*pow(i,base) else : num = num + (9 + ord(number[i])) *pow(i,base) return num a = baseencode('20',5) print a Errors I get are Traceback (most recent call last): File "doubtrob.py", line 19, in <module> a = baseencode('20',5) File "doubtrob.py", line 13, in baseencode if number[i]!= [A-Z]: NameError: global name 'A' is not defined

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  • im writing a spellchecking program, how do i replace ch in a string..eg..

    - by Ajay Hopkins
    what am i doing wrong/what can i do?? import sys import string def remove(file): punctuation = string.punctuation for ch in file: if len(ch) > 1: print('error - ch is larger than 1 --| {0} |--'.format(ch)) if ch in punctuation: ch = ' ' return ch else: return ch ref = (open("ref.txt","r")) test_file = (open("test.txt", "r")) dictionary = ref.read().split() file = test_file.read().lower() file = remove(file) print(file) p.s, this is in Python 3.1.2

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  • Detect if 2 HTML fragments have identical hierarchical structure

    - by sergzach
    An example of fragments that have identical hierarchical structure: (1) <div> <span>It's a message</span> </div> (2) <div> <span class='bold'>This is a new text</span> </div> An example of fragments that have different structure: (1) <div> <span><b>It's a message</b></span> </div> (2) <div> <span>This is a new text</span> </div> So, fragments with a similar structure correspond to one hierarchical tree (the same tag names, the same hierarchical structure). How can I detect if 2 elements (html fragments) have the same structure simply with lxml? I have a function that does not work properly for some more difficult case (than the example): def _is_equal( el1, el2 ): # input: 2 elements with possible equal structure and tag names # e.g. root = lxml.html.fromstring( buf ) # el1 = root[ 0 ] # el2 = root[ 1 ] # move from top to bottom, compare elements result = False if el1.tag == el2.tag: # has no children if len( el1 ) == len( el2 ): if len( el1 ) == 0: return True else: # iterate one of them, for example el1 i = 0 for child1 in el1: child2 = el2[ i ] is_equal2 = _is_equal( child1, child2 ) if not is_equal2: return False return True else: return False else: return False The code fails to detect that 2 divs with class='tovar2' have an identical structure: <body> <div class="tovar2"> <h2 class="new"> <a href="http://modnyedeti-krsk.ru/magazin/product/333193003"> ?????? ?/? </a> </h2> <ul class="art"> <li> ???????: <span>1759</span> </li> </ul> <div> <div class="wrap" style="width:180px;"> <div class="new"> <img src="shop_files/new-t.png" alt=""> </div> <a class="highslide" href="http://modnyedeti-krsk.ru/d/459730/d/820.jpg" onclick="return hs.expand(this)"> <img src="shop_files/fr_5.gif" style="background:url(/d/459730/d/548470803_5.jpg) 50% 50% no-repeat scroll;" alt="?????? ?/?" height="160" width="180"> </a> </div> </div> <form action="" onsubmit="return addProductForm(17094601,333193003,3150.00,this,false);"> <ul class="bott "> <li class="price">????:<br> <span> <b> 3 150 </b> ???. </span> </li> <li class="amount">???-??:<br><input class="number" onclick="this.select()" value="1" name="product_amount" type="text"> </li> <li class="buy"><input value="" type="submit"> </li> </ul> </form> </div> <div class="tovar2"> <h2 class="new"> <a href="http://modnyedeti-krsk.ru/magazin/product/333124803">?????? ?/?</a> </h2> <ul class="art"> <li> ???????: <span>1759</span> </li> </ul> <div> <div class="wrap" style="width:180px;"> <div class="new"> <img src="shop_files/new-t.png" alt=""> </div> <a class="highslide" href="http://modnyedeti-krsk.ru/d/459730/d/820.jpg" onclick="return hs.expand(this)"> <img src="shop_files/fr_5.gif" style="background:url(/d/459730/d/548470803_5.jpg) 50% 50% no-repeat scroll;" alt="?????? ?/?" height="160" width="180"> </a> </div> </div> <form action="" onsubmit="return addProductForm(17094601,333124803,3150.00,this,false);"> <ul class="bott "> <li class="price">????:<br> <span> <b>3 150</b> ???. </span> </li> <li class="amount">???-??:<br><input class="number" onclick="this.select()" value="1" name="product_amount" type="text"> </li> <li class="buy"> <input value="" type="submit"> </li> </ul> </form> </div> </body>

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  • os.walk in python not running with cmd line parameter passed as path

    - by kartiku
    Hello, I needed to find the number of files in a folder on the system. This is what i used: file_count = sum((len(f) for _, _, f in os.walk('path'))) This works fine when we specify the path as a string in quotes, but when I enter a variable name that holds the path, type(file_count) is a generator object, and hence cannot be used as an integer. How to solve this and why does this happen?

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  • Optimizing list comprehension to find pairs of co-prime numbers

    - by user3685422
    Given A,B print the number of pairs (a,b) such that GCD(a,b)=1 and 1<=a<=A and 1<=b<=B. Here is my answer: return len([(x,y) for x in range(1,A+1) for y in range(1,B+1) if gcd(x,y) == 1]) My answer works fine for small ranges but takes enough time if the range is increased. such as 1 <= A <= 10^5 1 <= B <= 10^5 is there a better way to write this or can this be optimized?

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  • Why won't haproxy capture my cookie?

    - by mike
    I'm having trouble getting frontend cookie capture to work in haproxy. I have this in my config: frontend frontend 0.0.0.0:9999 [snip] capture cookie foo len 10 Then I use nc to talk directly to the server and send it: GET / HTTP/1.1 Cookie: foo=bar I get a log line, but there's a "-" where the captured cookie should be.

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  • Use Twisted's getPage as urlopen?

    - by RadiantHex
    Hi folks, I would like to use Twisted non-blocking getPage method within a webapp, but it feels quite complicated to use such function compared to urlopen. This is an example of what I'm trying to achive: def web_request(request): response = urllib.urlopen('http://www.example.org') return HttpResponse(len(response.read())) Is it so hard to have something similar with getPage?

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  • Most concise way to convert from date format: yyyy[3 digit day of year] to SQL datetime

    - by Seth Reno
    I'm working with an existing database where all dates are stored as integers in the following format: yyyy[3 digit day of year]. For example: 2010-01-01 == 2010001 2010-12-31 == 2010356 I'm using the following SQL to convert to a datetime: DATEADD(d, CAST(SUBSTRING( CAST(NEW_BIZ_OBS_DATE AS VARCHAR), 5, LEN(NEW_BIZ_OBS_DATE) - 4 ) AS INT) - 1, CAST('1/1/' + SUBSTRING(CAST(NEW_BIZ_OBS_DATE AS VARCHAR),1,4) AS DATETIME)) Does anyone have a more concise way to do this?

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