NSString* testUrl = @"myprotocal://192.168.0.1:8080/test?username=user&password=user";
NSURL *myURL = [NSURL URLWithString:testUrl];
[[UIApplication sharedApplication] openURL:myURL ]
Category Archives: Language
Andoid调用Custom Protocol
String textUrl = "myprotocal://192.168.0.1:8080/test?username=user&password=user";
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(textUrl));
intent.addCategory(Intent. CATEGORY_BROWSABLE);
intent.addCategory(Intent. CATEGORY_DEFAULT);
startActivity(intent)
XCode空白工程，手工添加Storyboard步骤
XCode新建空白工程后，手工增加Storyboard并添加ViewController后，只显示空白，不显示ViewController：
调试时会提示：
Application windows are expected to have a root view controller at the end of application launch
这时，需要手工修改一处代码和一处配置：
1、AppDelegate源码文件
//修改前
- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
self.window = [[UIWindow alloc] initWithFrame:[[UIScreen mainScreen] bounds]];
// Override point for customization after application launch.
self.window.backgroundColor = [UIColor whiteColor];
[self.window makeKeyAndVisible];
return YES;
}
//修改后
- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
return YES;
}
2、指定启动的Storyboard
在Project-》Targets-》Summary中修改Main Storyboard属性为新添加的Storyboard名称（不带后缀）。
或者
修改XXX-Info.plist文件，添加属性
<key>UIMainStoryboardFile</key> <string>Storyboard名称（不带后缀）</string>
然后就OK啦。
汉字繁简体互换
1、chs2cht.h
#pragma once char* UnicodeToBIG5(const wchar_t* szUnicodeString); char* UnicodeToGB2312(const wchar_t* szUnicodeString); wchar_t* GB2312ToUnicode(const char* szGBString); char* GB2312ToBIG5(const char* szGBString);
2、chs2cht.cpp
#include "chs2cht.h"
#include <atlstr.h>
//GB2312 转 Unicode：
wchar_t* GB2312ToUnicode(const char* szGBString)
{
UINT nCodePage = 936; //GB2312
int nLength=MultiByteToWideChar(nCodePage,0,szGBString,-1,NULL,0);
wchar_t* pBuffer = new wchar_t[nLength+1];
MultiByteToWideChar(nCodePage,0,szGBString,-1,pBuffer,nLength);
pBuffer[nLength]=0;
return pBuffer;
}
//BIG5 转 Unicode：
wchar_t* BIG5ToUnicode(const char* szBIG5String)
{
UINT nCodePage = 950; //BIG5
int nLength=MultiByteToWideChar(nCodePage,0,szBIG5String,-1,NULL,0);
wchar_t* pBuffer = new wchar_t[nLength+1];
MultiByteToWideChar(nCodePage,0,szBIG5String,-1,pBuffer,nLength);
pBuffer[nLength]=0;
return pBuffer;
}
//Unicode 转 GB2312：
char* UnicodeToGB2312(const wchar_t* szUnicodeString)
{
UINT nCodePage = 936; //GB2312
int nLength=WideCharToMultiByte(nCodePage,0,szUnicodeString,-1,NULL,0,NULL,NULL);
char* pBuffer=new char[nLength+1];
WideCharToMultiByte(nCodePage,0,szUnicodeString,-1,pBuffer,nLength,NULL,NULL);
pBuffer[nLength]=0;
return pBuffer;
}
//Unicode 转 BIG5：
char* UnicodeToBIG5(const wchar_t* szUnicodeString)
{
UINT nCodePage = 950; //BIG5
int nLength=WideCharToMultiByte(nCodePage,0,szUnicodeString,-1,NULL,0,NULL,NULL);
char* pBuffer=new char[nLength+1];
WideCharToMultiByte(nCodePage,0,szUnicodeString,-1,pBuffer,nLength,NULL,NULL);
pBuffer[nLength]=0;
return pBuffer;
}
//繁体中文BIG5 转 简体中文GB2312
char* BIG5ToGB2312(const char* szBIG5String)
{
LCID lcid = MAKELCID(MAKELANGID(LANG_CHINESE,SUBLANG_CHINESE_SIMPLIFIED),SORT_CHINESE_PRC);
wchar_t* szUnicodeBuff = BIG5ToUnicode(szBIG5String);
char* szGB2312Buff = UnicodeToGB2312(szUnicodeBuff);
int nLength = LCMapStringA(lcid,LCMAP_SIMPLIFIED_CHINESE, (LPCSTR)szGB2312Buff,-1,NULL,0);
char* pBuffer = new char[nLength + 1];
LCMapStringA(0x0804,LCMAP_SIMPLIFIED_CHINESE,(LPCSTR)szGB2312Buff,-1,pBuffer,nLength);
pBuffer[nLength] = 0;
delete[] szUnicodeBuff;
delete[] szGB2312Buff;
return pBuffer;
}
//简体中文GB2312 转 繁体中文BIG5
char* GB2312ToBIG5(const char* szGBString)
{
LCID lcid = MAKELCID(MAKELANGID(LANG_CHINESE,SUBLANG_CHINESE_SIMPLIFIED),SORT_CHINESE_PRC);
int nLength = LCMapStringA(lcid,LCMAP_TRADITIONAL_CHINESE,szGBString,-1,NULL,0);
char* pBuffer=new char[nLength+1];
LCMapStringA(lcid,LCMAP_TRADITIONAL_CHINESE,szGBString,-1,pBuffer,nLength);
pBuffer[nLength]=0;
wchar_t* pUnicodeBuff = GB2312ToUnicode(pBuffer);
char* pBIG5Buff = UnicodeToBIG5(pUnicodeBuff);
delete[] pBuffer;
delete[] pUnicodeBuff;
return pBIG5Buff;
}
3、test.cpp
#include "chs2cht.h"
#include <iostream>
#include <locale.h>
#include <atlstr.h>
using namespace std;
int main(int argc, char** argv)
{
//“区域设置”为简体中文
locale loc( "chs" );
char str[100];
cin>>str;
char * rlt=GB2312ToBIG5(str);
CString cStr1;
cStr1.Format( TEXT("%s"),rlt);
//“区域设置”为繁体中文
setlocale(LC_ALL, ".950");
cout<<rlt<<endl;
return 0;
}
VC fix .pch file missing
表现：
Error 1 fatal error C1083: Cannot open precompiled header file: 'Debug\????.pch': No such file or directory
解决方法：
1、保证stdafx.h及stdafx.cpp在项目的最顶层，stdafx.h用于保存需要的头文件，stdafx.cpp中只有对stdafx.h的引用；
2、在VS中，右击需要修改的工程，选择”Properties”；
3、在左上角，选择“ All Configurations”；
4、在左边，找到“C/C++”，并定位到“Precompiled Headers”；
5、将选项Precompiled Header修改为： “Use (/Yu)”；
6、将选项“Precompiled Header File”修改为：“stdafx.h”；
7、保存设置；
8、保证#include “stdafx.h”为所有需要预编译的cpp文件的第一行；
9、VS中，右击stdafx.cpp，选择”Properties”；
10、在左上角，选择“ All Configurations”；
11、将选项Precompiled Header修改为： “Create (/Yc).”；
12、保存设置，重新编译。
JavaScript checking for null vs. undefined
JS中undefined表示变量没有定义或从来没有赋值，而null是空对象，是有值的。
例如：
var a; alert(typeof a); //"undefined" a = null; alert(typeof null); //"object"
JS中==比较为弱类型比较，JS会自动进行类型转换，然后返回值的比较结果。
而===为强类型比较，即必须类型与值同时相同，才会相等。
例如：
alert(0 == "0"); //true alert(0==="0"); //false alert(undefined == null); //true alert(undefined === null); //false
判断变量为null:
if (a === null) // or if (a == null)
第一种方法，当a为undefined时返回false，
第二种方法，当a为undefined时返回true。
判断变量为undefined:
if (typeof a === "undefined") // or if (a === undefined) // or if (a == undefined)
第一、二种方法，当a为null时返回false，
第三种方法，当a为null时返回true。
还有一种使用falsey进行检查的方法：
if (!a) {
// `a` is falsey, which includes `undefined` and `null`
// (and `""`, and `0`, and `false`)
}
大部分翻译自stackoverflow：
JavaScript checking for null vs. undefined and difference between == and ===
Prolog101(08)
列表：
放在用方括号中的一组项目的集合，各项目之间使用逗号分割。
空表：
没有项目的列表，用[]表示。
表头与表尾：
[X|Y]可以与任意的列表匹配，匹配成功后，X绑定为列表的第一个项目的值，我们称之为表头（head）。
而Y则绑定为剩下的列表，我们称之为表尾（tail）。
表尾(tail)一定是列表，而表头（head）则是一个项目，该项目可以是表，也可以是其他的任何数据结构。
看下这个例子就清楚了：
1 ?- [a|[b,c,d]] = [a,b,c,d]. true. 2 ?- [a|b,c,d] = [a,b,c,d]. ERROR: Syntax error: Unexpected comma or bar in rest of list ... 3 ?- [H|T] = [a]. H = a, T = []. 4 ?- [H|T] = [a,b,c,d]. H = a, T = [b, c, d]. 5 ?- [H|T] = [a,[b,c,d]]. H = a, T = [[b, c, d]]. 6 ?- [H|T] = []. false. 7 ?- [A,B|T] = [a,b,c,d]. A = a, B = b, T = [c, d]. 8 ?- [a|[b|[c|[d|[]]]]] = [a,b,c,d]. true.
检查数据是否存在：
1 ?- member(a, [a,b,c]). true . 2 ?- member(d, [a,b,c]). false. 3 ?- member(d, [a,b,c,[d],e]). false. 4 ?- member([d], [a,b,c,[d],e]). true . 5 ?- member(X, [a,b,c]). X = a ; X = b ; X = c.
追加
1 ?- append([a,b,c],[d,e,f],X). X = [a, b, c, d, e, f]. 2 ?- append([],[d,e,f],X). X = [d, e, f]. 3 ?- append([a,b],Y,[a,b,c,d]). Y = [c, d]. 4?- append(X,Y,[a,b,c]). X = [], Y = [a, b, c] ; X = [a], Y = [b, c] ; X = [a, b], Y = [c] ; X = [a, b, c], Y = [] ;
删除
1 ?- delete([a,b,c,d,e], c, X). X = [a, b, d, e]
Prolog101(07)
联合(Unification)
变量与任何项目: 变量可以与任何项目绑定，其中也包括变量
原始项目与原始项目: 两个原始项目（原子或整数）只有当它们相同时才能联合。
结构与结构: 如果两个结构的每个相应的参数能联合，那么这两个结构可以联合。
1 ?- (1,2,3)=(X,Y,Z). X = 1, Y = 2, Z = 3. 2 ?- (1,2,3)=(X,X,Z). false. 3 ?- (1,1,3)=(X,X,Z). X = 1, Z = 3. 4 ?- (1,X,3)=(X,1,Z). X = 1, Z = 3. 5 ?- (1,X,3)=(X,2,Z). false.
1 ?- (1,2,X)=(1,2,(3,4,5)). X = (3, 4, 5). 2 ?- (1,2,X)=(1,2,(3,4,Y)),Y=5. X = (3, 4, 5), Y = 5. %_表示不关心匹配内容 3 ?- (1,2,X,Y,7)=(1,2,(3,4,5),6,_). X = (3, 4, 5), Y = 6. 4 ?- X = Y, Y = hi, write(X). hi X = Y, Y = hi.
%swipl -s objs.pl
%Hansen
%room
room(kitchen).
%objects and location
%object(Name, Color, Size, Weight).
location(object(candle, red, small, 1), kitchen).
location(object(apple, red, small, 1), kitchen).
location(object(apple, green, small, 1), kitchen).
location(object(table, blue, big, 50), kitchen).
%current room
here(kitchen).
%can take something?
can_take(Thing) :-
here(Room),
location(object(Thing, _, small, _), Room).
can_take(Thing) :-
here(Room),
location(object(Thing, _, big, _), Room),
write('The '), write(Thing),
write(' is too big to carry.'), nl,
fail.
can_take(Thing) :-
here(Room),
not(location(object(Thing, _, _, _), Room)),
write('There is no '), write(Thing), write(' here.'), nl,
fail.
%out put the weight
write_weight(1) :- write('1 pound').
write_weight(W) :- W > 1, write(W), write(' pounds').
%list all things in a room
list_things(Place) :-
location(object(Thing, Color, Size, Weight), Place),
write('A '),write(Size),tab(1),
write(Color),tab(1),
write(Thing), write(', weighing '),
write_weight(Weight), nl,
fail.
Prolog101(06)
动态修改全局数据
asserta(X) :把子句X当作此子句的谓词的第一个子句加入到动态数据库中，不可回溯。
assert(X)或assertz(X) :把子句X当作此子句的谓词的最后一个子句加入到动态数据库中，不可回溯。
retract(X) :把子句X从动态数据库中删除，不可回溯。
%swipl -s room.pl
%Hansen
:-dynamic here/1.
:-dynamic location/2.
:-dynamic bag/1.
%房间定义
room(kitchen).
room(office).
room(hall).
room(diningroom).
room(cellar).
%门定义
door(office, hall).
door(kitchen, office).
door(hall, diningroom).
door(kitchen, cellar).
door(diningroom, kitchen).
%规则：有门的两个房间是相通的
connect(X,Y) :- door(X,Y).
connect(X,Y) :- door(Y,X).
%物品在哪个房间
location(desk, office).
location(apple, kitchen).
location(flashlight, desk).
location(washingmachine, cellar).
location(nani, washingmachine).
location(broccoli, kitchen).
location(crackers, kitchen).
location(computer, office).
%背包
bag(tourch).
%哪些物品可以吃
edible(apple).
edible(crackers).
edible(broccoli).
%当前位置
here(hall).
%移动
move(Place):- can_go(Place),retract(here(X)), asserta(here(Place)).
can_go(Place):- here(X), connect(X, Place).
can_go(Place):- write('You can''t go to '),write(Place),write(' from here.'), nl, fail.
%获取物品
take(X):- can_take(X), retract(location(X,_)), asserta(bag(X)), write(X), write(' taken.'), nl.
can_take(Thing):- here(Place), location(Thing, Place).
can_take(Thing):- write('There is no '), write(Thing), write(' here.'), nl, fail.
%放下物品
put(X):- can_put(X), here(Place), retract(bag(X)), asserta(location(X,Place)), write(X), write(' put.'), nl.
can_put(Thing):- bag(Thing).
can_put(Thing):- write('There is no '), write(Thing), write(' in your bag.'), nl, fail.
%房间物品列表
list_things(Place):- location(X, Place),tab(2),write(X),nl,fail.
list_things(_).
%相连的
list_connections(Place):- connect(Place, X),tab(2),write(X),nl,fail.
list_connections(_).
%持有物品列表
list_bag(Thing):- bag(X),tab(2),write(X),nl,fail.
list_bag(_).
%吃东西
eat(Thing):- can_eat(Thing), retract(bag(X)), write(Thing), write(' eaten. Yummy!').
can_eat(Thing):- bag(Thing), edible(Thing).
can_eat(Thing):- not(bag(Thing)), write('There is no '), write(Thing), write(' in your bag.'), nl, fail.
can_eat(Thing):- bag(Thing), write('You can''t eat the '), write(Thing), write('.'), nl, fail.
%查看房间情况
look :-
here(Place), write('You are in the '), write(Place), nl,
write('You can see:'),nl,list_things(Place),
write('You can go to:'), nl, list_connections(Place),
write('You have:'),nl,list_bag(Thing).
%帮助
game :-
write('Look around: look/0'),nl,
write('Move around: move/1'),nl,
write('Take something: take/1'),nl,
write('Eat something: eat/1').
1 ?- game. Look around: look/0 Move around: move/1 Take something: take/1 Eat something: eat/1 true. 2 ?- look. You are in the hall You can see: You can go to: diningroom office You have: tourch true. 3 ?- move(diningroom). true . 4 ?- look. You are in the diningroom You can see: You can go to: kitchen hall You have: tourch true. 5 ?- move(kitchen). true . 6 ?- look. You are in the kitchen You can see: apple broccoli crackers You can go to: office cellar diningroom You have: tourch true. 7 ?- take(apple). apple taken. true . 8 ?- look. You are in the kitchen You can see: broccoli crackers You can go to: office cellar diningroom You have: apple tourch true. 9 ?- eat(apple). apple eaten. Yummy! true .
Prolog101(05)
递归定义都包括两个部分：边界条件与递归部分。
边界条件定义最简单的情况。而递归部分，则首先解决一部分问题，然后再调用其自身来解决剩下的部分，
每一次都将进行边界检测，如果剩下的部分已经是边界条件中所定义的情况时，那么递归就圆满成功了。
递归解决阶乘问题
%swipl -s factorial.pl
%Hansen
fun(N):- N>0,fact(N,1).
fun(N):- write(N), write(' is smaller than 1. '),fail.
fact(1,M):- write(M),!.
fact(N,M):- N1 is N-1, M1 is M*N, fact(N1,M1).
1 ?- fun(0). 0 is smaller than 1. false. 2 ?- fun(1). 1 true . 3 ?- fun(2). 2 true . 4 ?- fun(3). 6 true . 5 ?- fun(4). 24 true . 6 ?- fun(5). 120 true . 7 ?- fun(6). 720 true .
递归解决汉诺塔问题
%swipl -s hanoi.pl
%Hansen
%N个盘子从A到C的问题，用递归解决的思路：
%N-1个盘子，从A到B
%1个盘子，从A到C
%N-1个盘子，从B到C
hanoi(N):-move(N,a,b,c).
move(1,A,_,C):-fromto(A,C),!.
move(N,A,B,C):-N1 is N-1,move(N1,A,C,B),fromto(A,C),move(N1,B,A,C).
fromto(Loc1,Loc2):-nl,write('move one disk from '),write(Loc1),write(' to '),write(Loc2).
执行查询：
1 ?- hanoi(1). move one disk from a to c true. 2 ?- hanoi(2). move one disk from a to b move one disk from a to c move one disk from b to c true. 3 ?- hanoi(3). move one disk from a to c move one disk from a to b move one disk from c to b move one disk from a to c move one disk from b to a move one disk from b to c move one disk from a to c true. %这里就会出错啦 4 ?- hanoi(0). ERROR: Out of local stack