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Search found 4 results on 1 pages for 'slf'.

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• Python easy_install confused on Mac OS X

  - by slf

  environment info: $ echo $PATH /opt/local/bin:/opt/local/sbin:/sw/bin:/sw/sbin:/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/usr/X11/bin:/usr/X11R6/bin:/opt/local/bin:/Developer/Platforms/iPhoneOS.platform/Developer/usr/bin:~/.utility_scripts $ which easy_install /usr/bin/easy_install specifically, let's try the simplejson module (I know it's the same thing as import json in 2.6, but that isn't the point) $ sudo easy_install simplejson Searching for simplejson Reading http://pypi.python.org/simple/simplejson/ Reading http://undefined.org/python/#simplejson Best match: simplejson 2.1.0 Downloading http://pypi.python.org/packages/source/s/simplejson/simplejson-2.1.0.tar.gz#md5=3ea565fd1216462162c6929b264cf365 Processing simplejson-2.1.0.tar.gz Running simplejson-2.1.0/setup.py -q bdist_egg --dist-dir /tmp/easy_install-Ojv_yS/simplejson-2.1.0/egg-dist-tmp-AypFWa The required version of setuptools (>=0.6c11) is not available, and can't be installed while this script is running. Please install a more recent version first, using 'easy_install -U setuptools'. (Currently using setuptools 0.6c9 (/System/Library/Frameworks/Python.framework/Versions/2.6/Extras/lib/python)) error: Setup script exited with 2 ok, so I'll update setuptools... $ sudo easy_install -U setuptools Searching for setuptools Reading http://pypi.python.org/simple/setuptools/ Best match: setuptools 0.6c11 Processing setuptools-0.6c11-py2.6.egg setuptools 0.6c11 is already the active version in easy-install.pth Installing easy_install script to /usr/local/bin Installing easy_install-2.6 script to /usr/local/bin Using /Library/Python/2.6/site-packages/setuptools-0.6c11-py2.6.egg Processing dependencies for setuptools Finished processing dependencies for setuptools I'm not going to speculate, but this could have been caused by any number of environment changes like the Leopard - Snow Leopard upgrade, MacPorts or Fink updates, or multiple Google App Engine updates.

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• Solving Combinatory Problems with LINQ /.NET4

  - by slf

  I saw this article pop-up in my MSDN RSS feed, and after reading through it, and the sourced article here I began to wonder about the solution. The rules are simple: Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number must also satisfy these divisibility requirements: The number should be divisible by 9. If the rightmost digit is removed, the remaining number should be divisible by 8. If the rightmost digit of the new number is removed, the remaining number should be divisible by 7. And so on, until there's only one digit (which will necessarily be divisible by 1). This is his proposed monster LINQ query: // C# and LINQ solution to the numeric problem presented in: // http://software.intel.com/en-us/blogs/2009/12/07/intel-parallel-studio-great-for-serial-code-too-episode-1/ int[] oneToNine = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; // the query var query = from i1 in oneToNine from i2 in oneToNine where i2 != i1 && (i1 * 10 + i2) % 2 == 0 from i3 in oneToNine where i3 != i2 && i3 != i1 && (i1 * 100 + i2 * 10 + i3) % 3 == 0 from i4 in oneToNine where i4 != i3 && i4 != i2 && i4 != i1 && (i1 * 1000 + i2 * 100 + i3 * 10 + i4) % 4 == 0 from i5 in oneToNine where i5 != i4 && i5 != i3 && i5 != i2 && i5 != i1 && (i1 * 10000 + i2 * 1000 + i3 * 100 + i4 * 10 + i5) % 5 == 0 from i6 in oneToNine where i6 != i5 && i6 != i4 && i6 != i3 && i6 != i2 && i6 != i1 && (i1 * 100000 + i2 * 10000 + i3 * 1000 + i4 * 100 + i5 * 10 + i6) % 6 == 0 from i7 in oneToNine where i7 != i6 && i7 != i5 && i7 != i4 && i7 != i3 && i7 != i2 && i7 != i1 && (i1 * 1000000 + i2 * 100000 + i3 * 10000 + i4 * 1000 + i5 * 100 + i6 * 10 + i7) % 7 == 0 from i8 in oneToNine where i8 != i7 && i8 != i6 && i8 != i5 && i8 != i4 && i8 != i3 && i8 != i2 && i8 != i1 && (i1 * 10000000 + i2 * 1000000 + i3 * 100000 + i4 * 10000 + i5 * 1000 + i6 * 100 + i7 * 10 + i8) % 8 == 0 from i9 in oneToNine where i9 != i8 && i9 != i7 && i9 != i6 && i9 != i5 && i9 != i4 && i9 != i3 && i9 != i2 && i9 != i1 let number = i1 * 100000000 + i2 * 10000000 + i3 * 1000000 + i4 * 100000 + i5 * 10000 + i6 * 1000 + i7 * 100 + i8 * 10 + i9 * 1 where number % 9 == 0 select number; // run it! foreach (int n in query) Console.WriteLine(n); Octavio states "Note that no attempt at all has been made to optimize the code", what I'd like to know is what if we DID attempt to optimize this code. Is this really the best this code can get? I'd like to know how we can do this best with .NET4, in particular doing as much in parallel as we possibly can. I'm not necessarily looking for an answer in pure LINQ, assume .NET4 in any form (managed c++, c#, etc all acceptable).

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• iPhone Game Developers - What does your toolchain look like?

  - by slf

  For example: source control: git + adobe drive 3d: google sketchup - *.dae - blender - *.obj 2d: photoshop/illustrator - *.png audio: audacity - *.caf code: ArgoUML, Xcode, Textmate test: OCUnit build: rake, Xcode Feel free to mention any other tools that you think are awesome :) Changed to Community Wiki

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• Lock-Free Data Structures in C++ Compare and Swap Routine

  - by slf

  In this paper: Lock-Free Data Structures (pdf) the following "Compare and Swap" fundamental is shown: template <class T> bool CAS(T* addr, T exp, T val) { if (*addr == exp) { *addr = val; return true; } return false; } And then says The entire procedure is atomic But how is that so? Is it not possible that some other actor could change the value of addr between the if and the assignment? In which case, assuming all code is using this CAS fundamental, it would be found the next time something "expected" it to be a certain way, and it wasn't. However, that doesn't change the fact that it could happen, in which case, is it still atomic? What about the other actor returning true, even when it's changes were overwritten by this actor? If that can't possibly happen, then why? I want to believe the author, so what am I missing here? I am thinking it must be obvious. My apologies in advance if this seems trivial.

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