cutepig's blog

谁能解释下这几行代码？谢谢了 [val,id]=sort(imgDiff(y,x,:)); if abs(val(1)-val(2))>10 dispMap(y,x)=id(1);

- 列表的匹配 15> [A|B]=[1,2,3,4]. [1,2,3,4] 16> A. 1 17> B. [2,3,4] 18> f(). ok 19> [A,B|C]=[1,2,3,4]. [1,2,3,4] 20> A. 1 21> B. 2 22> C. [3,4] 23> -模块 cutepig@ubuntu:~$ cat geometry.erl -module(geometry). -export([area/1]). area({rectangle,Width,Hit}) -> Width*Hit; area({circle,R}) -> 3.14159*R*R. cutepig@ubuntu:~$ erl Erlang R13B01 (erts-5.7.2) [source] [rq:1] [async-threads:0] [kernel-poll:false] Eshell V5.7.2 (abort with ^G) 1> c(geometry). {ok,geometry} 2> geometry:area({rectangle,10,5}). 50 3> geometry:area({circle,1.4}). 6.157516399999999 - 另一个例子 cutepig@ubuntu:~$ cat shop.erl -module(shop). -export([cost/1]). cost(oranges) -> 5; cost(newspaper) -> 8; cost(apples) -> 6. cutepig@ubuntu:~$ cat shop1.erl -module(shop1). -export([total/1]). total([{What,N}|T]) -> shop:cost(What)*N+total(T); total([]) -> 0. cutepig@ubuntu:~$ erl Erlang R13B01 (erts-5.7.2) [source] [rq:1] [async-threads:0] [kernel-poll:false] Eshell V5.7.2 (abort with ^G) 1> c(shop). {ok,shop} 2> c(shop1). {ok,shop1} 3> shop:cost(oranges). 5 4> shop:cost(orangesX). ** exception error: no function clause matching shop:cost(orangesX) 5> shop1:total([{oranges,2},{newspaper,3}]). 34 6> - 匿名函数 1> Z=fun(X) -> 2*X end. #Fun<erl_eval.6.13229925> 2> Z(2). 4 3> Double=Z. #Fun<erl_eval.6.13229925> 4> Double(3). 6 5> Hypot=fun(X,Y) -> math:sqrt(X*X+Y*Y) end. #Fun<erl_eval.12.113037538> 6> Hypot(3,4). 5.0 7> Hypot(3). ** exception error: interpreted function with arity 2 called with one argument 8> lists:map(Double,[1,2,3,4]). [2,4,6,8] 9> Even=fun(X) -> (X rem 2)==0 end. #Fun<erl_eval.6.13229925> 10> Even(2). true 11> Even(7). false 12> lists:map(Even,[1,2,3,4,5,6,7,8]). [false,true,false,true,false,true,false,true] 13> lists:filter(Even,[1,2,3,4,5,6,7,8]). [2,4,6,8] 多次定义的函数 14> TempConv=fun({c,C}) -> {f, 32+C*9/5}; 14> ({f,F}) -> {c, (F-32)*5/9} 14> end. #Fun<erl_eval.6.13229925> 15> TempConv({c,100}). {f,212.0} 16> TempConv({f,212}). {c,100.0} 高阶函数 17> MakeTest=fun(L) -> (fun(X) -> lists:member(X,L) end) end. #Fun<erl_eval.6.13229925> 18> IsFruit=MakeTest([apple,pear,orange]). #Fun<erl_eval.6.13229925> 19> IsFruit(apple). true 20> IsFruit(appleX). false 自己的for cutepig@ubuntu:~$ cat lib_misc. cat: lib_misc.: No such file or directory cutepig@ubuntu:~$ cat lib_misc.erl -module(lib_misc). -export([for/3]). for(Max,Max,F) -> [F(Max)]; for(I,Max,F) -> [F(I) | for(I+1,Max,F)]. 1> c(lib_misc). {ok,lib_misc} 2> lib_misc:for(1,10,fun(I) -> I end). [1,2,3,4,5,6,7,8,9,10] 3> lib_misc:for(1,10,fun(I) -> I*I end). [1,4,9,16,25,36,49,64,81,100] 自己的map和sum cutepig@ubuntu:~$ cat mylists.erl -module(mylists). -export([sum/1]). -export([map/2]). sum([H|T]) -> H + sum(T); sum([]) -> 0. map(_,[]) -> []; map(F,[H|T]) -> [F(H) | map(F,T)]. 1> c(mylists). {ok,mylists} 2> mylists:map(fun(X)->2*X end, [1,2,3]). [2,4,6] 3> 列表解析 1> L=[1,2,3,4]. [1,2,3,4] 2> [2*X || X <- L]. [2,4,6,8] cutepig@ubuntu:~$ cat mylists.erl -module(mylists). -export([sum/1]). -export([map/2]). -export([map2/2]). sum([H|T]) -> H + sum(T); sum([]) -> 0. map(_,[]) -> []; map(F,[H|T]) -> [F(H) | map(F,T)]. map2(F,L) -> [F(X) || X <- L]. 1> c(mylists). {ok,mylists} 2> mylists:map(fun(X) -> X*2 end, [1,2,3]). [2,4,6] 快速排序 cutepig@ubuntu:~$ cat lib_misc.erl -module(lib_misc). -export([for/3]). -export([qsort/1]). for(Max,Max,F) -> [F(Max)]; for(I,Max,F) -> [F(I) | for(I+1,Max,F)]. qsort([]) -> []; qsort([Pivot|T]) -> qsort([X || X <- T, X < Pivot]) ++ [Pivot] ++ qsort([X || X <- T, X >= Pivot]). 1> c(lib_misc). {ok,lib_misc} 2> lib_misc:qsort([3,2,4,1]). [1,2,3,4]

http://stackoverflow.com/questions/2544430/python-mechanize-browser-submit-related-problem mechanize doesn't support Javascript at all. If you absolutely have to run that Javascript, look into Selenium. It offers python bindings to control a real, running browser like Firefox or IE.

example: http://stackoverflow.com/questions/6369097/http-error-405-500-in-python-mechanize-using-mechanize-auto-login-one-website username = 'staryin' password = 'zxy1985' login_url = 'http://www.autohome.com.cn/tools/hometoplogin.html' br = Browser() cj = cookielib.LWPCookieJar() br.set_cookiejar(cj) br.addheaders = [('User-agent', 'Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.9.0.1) Gecko/2008071615 Fedora/3.0.1-1.fc9 Firefox/3.0.1')] br.open(login_url) br.select_form(nr = 0) br['name'] = username br['pwd'] = password br.form.action = login_url br.submit().read()

这个博客很帅

我就搜一下list的用法， 居然搞到电影list。。。 囧啊。。。

网上关于WTL的文章好像不多，非常感谢您能发布这些资料，但是为什么会在全文打包下载时无法下载并产生404错误呢？

萨达是发生的水电费圣代

阿双方的身份的说法舒服点

happy accident,想不到搜list竟然搜到了电影

感谢分享！

没有明白代码Select的思想，求详细些的解释。

黑书上的题目都很经典啊！这么多年也不会忘记

这个有难度啊

不错

http://www.debuginfo.com/articles/easywindbg.html # Solving real life problems * Debugging deadlocks !locks ~*kpn * Debugging high CPU utilization !runaway ~*kpn * Debugging stack overflow ~*kf; * Saving dumps .dump /m c:\myapp.dmp * Crash dump analysis cdb -z c:\myapp.dmp -logo out.txt -lines -c "!analyze -v;q" * Virtual memory analysis !vadump -v; !address; * Searching for symbols x Module!Symbol ln 0x77d491c8 * Displaying data structures dt -b CSymbolInfoPackage

请问你有A Fast Area-Based Stereo Matching Algorithm这篇文章的中文翻译吗

@陈梓瀚(vczh) 呀，大名鼎鼎的vczh拜访了呀，还没怎么看呢

MAC 地址： 00-23-CD-8B-D0-97 IP地址： 116.48.50.150 PPPoE 子网掩码： 255.255.255.255 网关： 116.48.50.150 DNS 服务器： 219.76.98.66, 218.102.32.208 上网时间： 0 day 08:23:32

我很想知道你分析成什么样子了……

LuaWrapper的地址已经无法下载了

准确的说着不是用CPropertySheet, CPropertyPage实现的标签对话框

[code=cpp] struct X { int foo() { return 0; } } x; struct Y { static int (X::*p)(); }; int (X::*Y::p)()=&X::foo; [/code] codes compiled error.

// testBoost.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include <boost/fusion/sequence.hpp> #include <boost/fusion/include/sequence.hpp> #include <boost/fusion/include/vector.hpp> #include <boost/fusion/include/map.hpp> #include <string> #include <iostream> #include <boost/fusion/algorithm.hpp> #include <boost/fusion/include/algorithm.hpp> #include <boost/fusion/support/is_sequence.hpp> #include <boost/fusion/include/is_sequence.hpp> struct print_xml { template< typename T > void operator()(T const& x) const { print_(x, boost::fusion::traits::is_sequence<T>() ); } private: // x is not sequence template <typename T> void print_(T const& x, const boost::mpl::false_&) const { std::cout << '<' << typeid(x).name() << '>' << x << "\n" ; } // x is sequence template <typename T> void print_(T const& x, const boost::mpl::true_&) const { boost::fusion::for_each(x, print_xml()); } }; int _tmain(int argc, _TCHAR* argv[]) { typedef boost::fusion::vector<int, char, std::string> Stuff; Stuff stuff(1, 'x', "howdy"); boost::fusion::vector<int, char, Stuff> stuff2(2, 'y', stuff); int i = boost::fusion::at_c<0>(stuff); char ch = boost::fusion::at_c<1>(stuff); std::string s = boost::fusion::at_c<2>(stuff); //boost::fusion::for_each(stuff, print_xml()); print_xml()(stuff2); return 0; }

請問先進, 我看您說明您的思路是 --------------- .找特征： 对上一帧拍摄的左右图像分别计算SIFT特征，保存为L1.key,R1.key。 对当前帧拍摄的左右图像分别计算SIFT特征，保存为L2.key,R2.key。 --------------- 但是怎麼會下四次找keypoint? 如果在單張影像中就可以找keypoint, 那思路是? 另外, 你這些程式(命令), 除了siftWin32.exe外, match就是自己寫的啊? 還是在matlab內執行? 可以說明一下keypoint的格式嗎? 也可以說看看*.match的格式嗎? --------------- >siftWin32.exe 0<L1.pgm 1>L1.key Finding keypoints 2473 keypoints found. >siftWin32.exe 0<R1.pgm 1>R1.key Finding keypoints 2945 keypoints found. >siftWin32.exe 0<L2.pgm 1>L2.key Finding keypoints 2712 keypoints found. >siftWin32.exe 0<R2.pgm 1>R2.key Finding keypoints 3536 keypoints found.

tolua http://www.codenix.com/~tolua/#documentation

http://67.223.234.84/boost_doc/doc/html/bbv2.html

vs2008编译不了driverStudio里面的代码, 而我下载的winddk又不支持vc6? VS2008可以支持DDK的。 或者VC+DDK

終於明白code footprint是什麽意思了，還是谷哥信得過。

你计算出来的dc根本就不是dark channel吧？！ 我觉得不是，效果根本就不一样，与原文计算出来的dark channel。因为，你没有采用window来做。 期待你的回答。 期待跟你交流。

ddd后者kde界面的kdbg DDD 或者 insight 都很小巧，ddd就是gdb的图形版

楼主有Yi Ma , Stefano Soatto. An Invitation to 3-D Vision , From Images to Geometric Models 这本书的电子版吗？能否发我一份： simba.forrest@gmail.com 谢谢啦！

Lets take an example: Class B { ... int iMem1; int iMem2; } Class D: public B { ... int iMem; } In this case, sizeof(D) is will also include the size of B. So it will be 12 bytes. The existence of virtual function(s) Existence of virtual function(s) will add 4 bytes of virtual table pointer in the class, which will be added to size of class. Again, in this case, if the base class of the class already has virtual function(s) either directly or through its base class, then this additional virtual function won't add anything to the size of the class. Virtual table pointer will be common across the class hierarchy. That is class Base { public: ... virtual void SomeFunction(...); private: int iAMem }; class Derived : public Base { ... virtual void SomeOtherFunction(...); private: int iBMem }; In the example above, sizeof(Base) will be 8 bytes--that is sizeof(int iAMem) + sizeof(vptr). sizeof(Derived) will be 12 bytes, that is sizeof(int iBMem) + sizeof(Derived). Notice that the existence of virtual functions in class Derived won't add anything more. Now Derived will set the vptr to its own virtual function table. Compiler being used In some scenarios, the size of a class object can be compiler specific. Lets take one example: class BaseClass { int a; char c; }; class DerivedClass : public BaseClass { char d; int i; }; If compiled with the Microsoft C++ compiler, the size of DerivedClass is 16 bytes. If compiled with gcc (either c++ or g++), size of DerivedClass is 12 bytes. The reason for sizeof(DerivedClass) being 16 bytes in MC++ is that it starts each class with a 4 byte aligned address so that accessing the member of that class will be easy (again, the memory read/write cycle). Mode of inheritance (virtual inheritance) In C++, sometimes we have to use virtual inheritance for some reasons. (One classic example is the implementation of final class in C++.) When we use virtual inheritance, there will be the overhead of 4 bytes for a virtual base class pointer in that class. class ABase{ int iMem; }; class BBase : public virtual ABase { int iMem; }; class CBase : public virtual ABase { int iMem; }; class ABCDerived : public BBase, public CBase { int iMem; }; And if you check the size of these classes, it will be: * Size of ABase : 4 * Size of BBase : 12 * Size of CBase : 12 * Size of ABCDerived : 24 Because BBase and CBase are dervied from ABase virtually, they will also have an virtual base pointer. So, 4 bytes will be added to the size of the class (BBase and CBase). That is sizeof ABase + size of int + sizeof Virtual Base pointer. Size of ABCDerived will be 24 (not 28 = sizeof (BBase + CBase + int member)) because it will maintain only one Virtual Base pointer (Same way of maintaining virtual table pointer).

Determining the Size of a Class Object By Girish Shetty There are many factors that decide the size of an object of a class in C++. These factors are: 1. Size of all non-static data members 2. Order of data members 3. Byte alignment or byte padding 4. Size of its immediate base class 5. The existence of virtual function(s) (Dynamic polymorphism using virtual functions). 6. Compiler being used 7. Mode of inheritance (virtual inheritance) Size of all non-static data members Only non-static data members will be counted for calculating sizeof class/object. class A { private: float iMem1; const int iMem2; static int iMem3; char iMem4; }; For an object of class A, the size will be the size of float iMem1 + size of int iMem2 + size of char iMem3. Static members are really not part of the class object. They won't be included in object's layout. <2>Order of data members The order in which one specifies data members also alters the size of the class. class C { char c; int int1; int int2; int i; long l; short s; }; The size of this class is 24 bytes. Even though char c will consume only 1 byte, 4 bytes will be allocated for it, and the remaining 3 bytes will be wasted (holes). This is because the next member is an int, which takes 4 bytes. If we don't go to the next (4th) byte for storing this integer member, the memory access/modify cycle for this integer will be 2 read cycles. So the compiler will do this for us, unless we specify some byte padding/packing. If I re-write the above class in different order, keeping all my data members like below: class C { int int1; int int2; int i; long l; short s; char c; }; Now the size of this class is 20 bytes. In this case, it is storing c, the char, in one of the slots in the hole in the extra four bytes. Byte alignment or byte padding As mentioned above, if we specify 1 byte alignment, the size of the class above (class C) will be 19 in both cases. Size of its immediate base class The size of a class also includes size of its immediate base class.

概念: 词法分析 lexical analysis 正则表达式 regular expression 非确定有限自动机 NFA (non-deterministic Finite Automata) 确定有限自动机 DFA 语法分析 syntax analysis, 利用LR等分析方法检查语法是否有问题 上下文无关文法 Context-free grammar 产生式 production 符号symbol 终结符 terminal 非终结符nonterminal 开始符号 start symbol 推导 derivation 最左推导 leftmost derivation 最右推导 rightmost derivation 语法分析树parse tree 二义性的ambiguous 左结合Left binding 递归下降 recursive descending first集合, follow集合 左递归 Left recursion LR分析 移近shift (转换goto) 规约reduce 接受 accept closure(I), goto(I,X) , 项集合 SLR LALR 语义分析 Semantic Analysis, 检查变量类型/函数参数/返回值 yacc文件结构 %{ ... %} 定义token的不同数据类型 %union { int iValue; /* integer value */ char sIndex; /* symbol table index */ nodeType *nPtr; /* node pointer */ }; %token <iValue> INTEGER %token <sIndex> VARIABLE %token WHILE IF PRINT FOR TO FUNCTION FUNCTION_DECL FUNCTION_CALL FUNCTION_IMPL ARGUMENT_LIST RETURN 下面这个用来通过修改优先级,避免冲突 %nonassoc IFX %nonassoc ELSE 左结合 %left GE LE EQ NE '>' '<' %left '+' '-' %left '*' '/' %nonassoc UMINUS 文法中项$x的类型 %type <nPtr> stmt expr stmt_list function_call function_impl function_decl argument_list %% program: stmt_list { ex($1); } | xx ; 错误恢复

paint.net 不开源了

[quote]smilingcat：博主如果窗口大小调整一下，效果应该会好一些，论文使用的大小是15*15，博主的可能有点小了，但是我试验了一下，论文方法的确对于这张图片的效果不是很好。另外有一点想请教博主，关于image matting这一步骤，能讲解一下吗？我没有看明白。[/quote] image matting我没看, 所以也不懂, 囧~~~

[quote]westlifegc：minfilt2函数的内容呢。[/quote] @westlifegc minfilt2在这里 http://www.mathworks.com/matlabcentral/fileexchange/1358

博主如果窗口大小调整一下，效果应该会好一些，论文使用的大小是15*15，博主的可能有点小了，但是我试验了一下，论文方法的确对于这张图片的效果不是很好。另外有一点想请教博主，关于image matting这一步骤，能讲解一下吗？我没有看明白。

minfilt2函数的内容呢。

不错，好文章，长知识了

分析 画每一行,包括画左边那个可以点的+号, 中间那个垂直分割线 LONG CPropTreeItem::DrawItem(CDC* pDC, const RECT& rc, LONG x, LONG y) 点击+号变鼠标光标 (那个垂直分割线地实现也是这样) BOOL CPropTreeList::OnSetCursor(CWnd* pWnd, UINT nHitTest, UINT message) 点击+号可以expand列表 (那个垂直分割线地实现也是这样) void CPropTreeList::OnLButtonDown(UINT, CPoint point) 这里根据点击的位置,有几种可能 HTCOLUMN 拖动垂直分割线 HTCHECKBOX点击左边column的check box HTEXPAND点击左边的那个+号 HTATTRIBUTE: 一般是点击右边的自定义控件区域 HTLABEL ?? HTCLIENT 几个类的关系 实现自己的自定义类

C++啊,以为是C#呢!

博客园里也谈文学？不过说的是不错的

up

@阿熊 仔细看了看,是我写的错了~

你好， 我认为你程序for循环中的:[Xp,P]=Correct(P,H,R,X,z_m); 应该改为:[Xp,P]=Correct(P,H,R,Xp,z_m); 改了后效果不如你的贴图。根据相关文献以及上面的程序实例，修改后的才是正确的吧。请大家讨论讨论我说的是否正确！

Java 程序员面试论坛 http://www.222345.net

Arnodi可能是比较合适的..但是没有挖掘稀疏性

@SmartPtr
那该怎样在文件开始定义泛化版本呢？
请明士，3ks。