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• Setting up SANED with sane-test

  - by Enoon

  I'm trying to setup a network service for running saned (running the sane-test backend) on Ubuntu 12.10, running it on a virtualbox. I followed the directions found here: https://help.ubuntu.com/community/ScanningHowTo and i got to the point where if i use a front-end like xsane, or use the command scanimage -d test i get the desired behaviour. (i get the test image). But when i try to use the network demon to access the backend (from the localhost) i get Failed to connect. I used telnet on 127.0.0.1 6566 and i got the following error: saned: symbol lookup error: saned: undefined symbol: sanei_w_init I'm a linux newbie, so i could be making some stupid mistake. Any ideas on how to fix this? Thanks [update]After posting this question i logged out, logged back in and it worked. After a few days i tried again and it gave the same error as before. Any help?

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• Pattern for loading and handling resources

  - by Enoon

  Many times there is the need to load external resources into the program, may they be graphics, audio samples or text strings. Is there a patten for handling the loading and the handling of such resources? For example: should I have a class that loads all the data and then call it everytime I need the data? As in: GraphicsHandler.instance().loadAllData() ...//and then later: draw(x,y, GraphicsHandler.instance().getData(WATER_IMAGE)) //or maybe draw(x,y, GraphicsHandler.instance().WATER_IMAGE) Or should I assign each resource to the class where it belongs? As in (for example, in a game): Graphics g = GraphicsLoader.load(CHAR01); Character c = new Character(..., g); ... c.draw(); Generally speaking which of these two is the more robust solution? GraphicsHandler.instance().getData(WATER_IMAGE) //or GraphicsHandler.instance().WATER_IMAGE //a constant reference

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• Removing duplicate solutions

  - by Enoon

  My code merges two lists of lists, item by item, in the following way: mergeL([[a,b],[c,d]], [[1,2],[3,4]], Result). Result = [[a,b,1,2],[c,d,3,4]] And this is the code i use: mergeL([],[],[]). mergeL(List, [], List). mergeL([], List, List). mergeL([X|Rest],[Y|Rest2], [XY|Res2]) :- mergeL(Rest, Rest2, Res2), append(X,Y,XY). This seems to work but if i call it with two lists of the same size i get three repeated results. Example (both list contain only one element): ?- mergeL([[a,b]],[[1,2,3]],Q). Q = [[a, b, 1, 2, 3]] ; Q = [[a, b, 1, 2, 3]] ; Q = [[a, b, 1, 2, 3]]. Is there a clean way to make this output only one solution?

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