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RSS as an injection vector
Blog feeds may carry security risk Several months ago, I myself independently found that apps relying on RSS (or ATOM or whatever) can be extremely vulnerable if subscribers/app developers do not pay much attention to the XSS attack. I believe many meta blogs and RSS readers are vulnerable to those type of attacks. But actual…
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Human Computation
Google TechTalks July 26, 2006 The talk given at Google on Human Computation. Luis von Ahn from CMU discuss how to encourage people to participate in Human Computation and gain fun and profit from them. He speaks very fast but very clearly and concisely, and I like that. I myself played a couple of games…
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Can Your Programming Language Do This?
Can Your Programming Language Do This? 그래서, 당신이 좋아하는 언어 X에서는 이게 되나요? 주로 Map, reduce, join 같은 것들을 이야기함. 그것들만이 functional language의 본질인 것 처럼 말하는 것은 다소이상해보임. 오히려 side effect를 없애는 모듈화 및 함수를 통한 문제 정의가 핵심이 아니었던가.
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template class? class template?
http://www.comeaucomputing.com/techtalk/templates/#terms 1998년 표준까지는 혼용하여 사용하다 2003년에 구분되기 시작함.
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compiz
So, what is compiz? Seeing is believing: I’ve succeeded in installing compiz. It’s very amazing and convenient. I’m sure that compiz is a must see killer app. But installation is PAIN!!! You need a couple of all nights especially if you have ATI. (And I do have ATI. Sad.) Compiz should be installed in the…
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@Override for method overriding
If I compile this:
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The Continuing Adventures of Java Puzzlers
The Continuing Adventures of Java Puzzlers : Tiger Traps : TS-1188, 2006 Finally, Java Puzzler for Tiger! Definitely one of the best on Java Core. The presentation is full of fun and pun; for example, bleep and blat.
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What’s New in Java SE 6 Beta 2 (Mustang)
What’s New in Java SE 6 Beta 2 (Mustang)
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memcpy로 vector복사
으 정말 끝장;;;
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Inlining C code in Ruby
http://on-ruby.blogspot.com/2006/07/rubyinline-making-making-things-faster.html C 코드를 Ruby에 inlining하는 방법입니다. 먼저 inline gem 설치. 테스트 코드. require “rubygems” require “inline” class RubyInlinePrime inline do |builder| builder.c ” int prime(int n) { int x; if (n == 2) return 1; for (x = 2; x < n - 1; x++) if (n % x == 0) return x; return 1;…
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