类型转换

类型转换可以判断实例的类型，也可以将实例看做是其父类或者子类的实例。

类型转换在Swift中使用is和as操作符实现。这两个操作符提供了一种简单达意的方式去检查值的类型或者转换它的类型。

你也可以用它来检查一个类型是否实现了某个协议。

定义一个类层次作为例子

可以将类型转换用在类和子类的层次结构上，检查特定类实例的类型并且转换这个类实例的类型成为这个层次结构中的其他类型。

/** * 这个类为任何出现在数字媒体库的媒体项提供基础功能 */
class MediaItem {
    var name: String
    init(name: String) {
        self.name = name
    }
}

class Movie: MediaItem {
    var director: String
    init(name: String,director: String) {
        self.director = director
        super.init(name: name)
    }
}

class Song: MediaItem {
    var artist: String
    init(name: String,artist: String) {
        self.artist = artist
        super.init(name: name)
    }
}

let library = [     // 类型被推断为[MediaItem]
    Movie(name: "Casablanca",director: "Michael Curtiz"),Song(name: "Blue Suede Shoes",artist: "Elvis Presley"),Movie(name: "Citizen Kane",director: "Orson Welles"),Song(name: "The One And Only",artist: "Chesney Hawkes"),Song(name: "Never Gonna Give You Up",artist: "Rick Astley")
]
print(library)      // [Movie,Song,Movie,Song]

推断出[MediaItem]类作为library的类型。在幕后library里存储的媒体项依然是Movie和Song类型的。但是，若你迭代它，依次取出的实例会是MediaItem类型的，而不是Movie和Song类型。为了让它们作为原本的类型工作，你需要检查它们的类型或者向下转换它们到其它类型。

检查类型

用类型检查操作符（is）来检查一个实例是否属于特定子类型。若实例属于那个子类型，类型检查操作符返回true，否则返回false。

/** * 这个类为任何出现在数字媒体库的媒体项提供基础功能 */
class MediaItem {
    var name: String
    init(name: String) {
        self.name = name
    }
}

class Movie: MediaItem {
    var director: String
    init(name: String,Song]

var movieCount = 0
var songCount = 0

for i in library {
    if i is Movie {
        ++movieCount
    } else if i is Song {
        ++songCount
    }
}
print("\(movieCount) movies,\(songCount) songs")   // 2 movies,3 songs

向下转型

某类型的一个常量或变量可能在幕后实际上属于一个子类。当确定是这种情况时，你可以尝试向下转到它的子类型，用类型转换操作符（as?或as!）。

因为向下转型可能会失败，类型转型操作符带有两种不同形式。条件形式（conditional form）as?返回一个你试图向下转成的类型的可选值（optional value）。强制形式``as! 把试图向下转型和强制解包（force-unwraps）转换结果结合为一个操作。

当你不确定向下转型可以成功时，用类型转换的条件形式（as?）。条件形式的类型转换总是返回一个可选值（optional value），并且若下转是不可能的，可选值将是nil。这使你能够检查向下转型是否成功。

只有你可以确定向下转型一定会成功时，才使用强制形式（as!）。当你试图向下转型为一个不正确的类型时，强制形式的类型转换会触发一个运行时错误。

转换没有真的改变实例或它的值。根本的实例保持不变；只是简单地把它作为它被转换成的类型来使用。

/** * 这个类为任何出现在数字媒体库的媒体项提供基础功能 */
class MediaItem {
    var name: String
    init(name: String) {
        self.name = name
    }
}

class Movie: MediaItem {
    var director: String
    init(name: String,Song]

for i in library {
    if let item = i as? Movie {
        print("Movie: name = \(item.name),director = \(item.director)")
    } else if let item = i as? Song {
        print("Song: name = \(item.name),artist = \(item.artist)")
    }
}
/* Movie: name = Casablanca,director = Michael Curtiz Song: name = Blue Suede Shoes,artist = Elvis Presley Movie: name = Citizen Kane,director = Orson Welles Song: name = The One And Only,artist = Chesney Hawkes Song: name = Never Gonna Give You Up,artist = Rick Astley */

Any和AnyObject的类型转换

Swift为不确定类型提供了两种特殊的类型别名：

• AnyObject可以表示任何类类型的实例
• Any可以表示任何类型，包括函数类型。

注意，只有当你确实需要它们的行为和功能时才使用Any和AnyObject。在你的代码里使用你期望的明确类型总是更好的。

AnyObject类型

当在工作中使用Cocoa APIs时，我们经常会接收到一个[AnyObject]类型的数组，或者说“一个任意类型对象的数组”。这是因为Objective-C没有明确的类型化数组。但是，你常常可以从API提供的信息来确定数组中对象的类型。

译者注
这段文档似乎没有及时更新，从Xcode 7和Swift 2.0开始，由于Objective-C引入了轻量泛型，集合类型已经可以类型化了，在Swift中使用Cocoa API也越来越少遇到AnyObject类型了。详情请参阅Lightweight Generics和Collection Classes。

在这些情况下，你可以使用强制形式的类型转换（as）来下转数组中的每一项到比AnyObject更明确的类型，不需要可选解包（optional unwrapping）。

/** * 这个类为任何出现在数字媒体库的媒体项提供基础功能 */
class MediaItem {
    var name: String
    init(name: String) {
        self.name = name
    }
}

class Movie: MediaItem {
    var director: String
    init(name: String,artist: String) {
        self.artist = artist
        super.init(name: name)
    }
}

let library: [AnyObject] = [
    Movie(name: "2001: A Space Odyssey",director: "Stanley Kubrick"),Movie(name: "Moon",director: "Duncan Jones"),Movie(name: "Alien",director: "Ridley Scott")
]
print(library)      // [Movie,Movie]

// 因为知道这个数组只包含Movie 实例，你可以直接用（as!）下转并解包到非可选的Movie类型
for item in library {
    let movie = item as! Movie
    print("Movie: name = \(movie.name),director = \(movie.director)")
}
print("==========")
// 为了变为一个更简短的形式，下转数组为[Movie]类型而不是下转数组中的每一项
for movie in library as! [Movie] {
    print("Movie: name = \(movie.name),director = \(movie.director)")
}
/* Movie: name = 2001: A Space Odyssey,director = Stanley Kubrick Movie: name = Moon,director = Duncan Jones Movie: name = Alien,director = Ridley Scott ========== Movie: name = 2001: A Space Odyssey,director = Ridley Scott */

在上面的代码中，可以下转数组中的各项，也可以直接下转整个数组，前提是实际类型是相同的。

Any类型

使用Any类型来和混合的不同类型一起工作，包括函数类型和非类类型。

/** * 这个类为任何出现在数字媒体库的媒体项提供基础功能 */
class MediaItem {
    var name: String
    init(name: String) {
        self.name = name
    }
}

class Movie: MediaItem {
    var director: String
    init(name: String,director: String) {
        self.director = director
        super.init(name: name)
    }
}

var things = [Any]()
things.append(0)
things.append(0.0)
things.append(42)
things.append(3.14159)
things.append("hello")
things.append((3.0,5.0))
things.append(Movie(name: "Ghostbusters",director: "Ivan Reitman"))
things.append({ (name: String) -> String in "Hello,\(name)" })

for thing in things {
    switch thing {
    case 0 as Int:    // switch语句的case绑定它们匹配到的值到一个指定类型的常量
        print("0 as Int,thing = \(thing)")
    case 0 as Double:    // switch语句的case绑定它们匹配到的值到一个指定类型的常量
        print("0 as Double,thing = \(thing)")
    case let i as Int:
        print("Int: \(i)")
    case let d as Double where d > 0:
        print("positive double = \(d),thing = \(thing)")
    case is Double:
        print("negative double,thing = \(thing)")
    case let s as String:
        print("string = \(s)")
    case let (x,y) as (Double,Double):
        print("double tuple = (\(x),\(y))")
    case let movie as Movie:
        print("movie.name = \(movie.name),director = \(movie.director)")
    case let c as String -> String:
        print("a closure: " + c("Tim"))
    default:
        print("other...,thing = \(thing)")
    }
}
/* 0 as Int,thing = 0 0 as Double,thing = 0.0 Int: 42 positive double = 3.14159,thing = 3.14159 string = hello double tuple = (3.0,5.0) movie.name = Ghostbusters,director = Ivan Reitman a closure: Hello,Tim */

#include"stdio.h"
#include"string.h"
#include"queue"
#include"vector"
#include"algorithm"
#include"iostream"
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=100010;//点数 
const int maxm=400010;//边数
struct node{
	int v,next,cap,flow;
}edge[maxm];
int cnt;
int head[maxn];
int cur[maxn],d[maxn];// 当前弧下标   结点到汇点弧长 
int p[maxn],gap[maxn];////可增广路上的上一条弧   gap优化
int ss[maxn];//保存路径
void init(){
	cnt=-1;
	memset(head,-1,sizeof(head));
} 
void debug(int k){
	int i,j;
	printf("\n");
	for (i=0;i<=k;i++) 
	for (j=head[i];j!=-1;j=edge[j].next) 
        
        printf("%d %d %d\n",i,edge[j].v,edge[j].flow);
}
void add(int u,int v,int w,int rw=0){
	cnt++;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	edge[cnt].cap=w;
	edge[cnt].flow=0;
	head[u]=cnt;
	cnt++;
	edge[cnt].v=u;
	edge[cnt].next=head[v];
	edge[cnt].cap=rw;
	edge[cnt].flow=0;
	head[v]=cnt;
}
void bfs(int k){
	int v,i;
	int u;
	memset(gap,0,sizeof(gap));
	memset(d,-1,sizeof(d));
	d[k]=0;
	gap[0]++;
	queue<int> q;
	q.push(k);
	while(q.empty()==false){
		u=q.front();q.pop();
		for (i=head[u];i!=-1;i=edge[i].next){
			v=edge[i].v;
			if (d[v]==-1) {
				d[v]=d[u]+1;
				gap[d[v]]++;
				q.push(v);
			}
		}
	}
}
int sap(int s,int t,int N){
	int i;
	bfs(t);
	memcpy(cur,head,sizeof(head));
	int top=0;
	int u=s;
	int ans=0;
	while(d[s]<N){
		if (u==t){
			int min=inf;
			int inser;
			for (i=0;i<top;i++){
				if (min>=edge[ss[i]].cap-edge[ss[i]].flow){
					min=edge[ss[i]].cap-edge[ss[i]].flow;
					inser=i;//这跟管子是满流了 
				}
			}
			for (i=0;i<top;i++){
				edge[ss[i]].flow+=min;
				edge[ss[i]^1].flow-=min;
			} 
			ans+=min;
			top=inser;
			u=edge[ss[top]^1].v;//u为满流的那个结点 也就是那根管子的倒管子的终点
			continue;
		}
		bool ok=false;
		int v;
		for (i=cur[u];i!=-1;i=edge[i].next){ 
			v=edge[i].v;
			if (edge[i].cap-edge[i].flow>0&&d[v]+1==d[u]){
				ok=true;
				cur[u]=i;
				break;
			}//u后 添加了一条下标为i的弧 
		}
		if(ok==true){
			ss[top]=cur[u];
			top++;
			u=v;
			continue;
		}
		//如果这条路已经走不通了，那么撤退
		int min=N;
		for (i=head[u];i!=-1;i=edge[i].next)
		  if (edge[i].cap-edge[i].flow>0&&d[edge[i].v]<min){
		  	min=d[edge[i].v];
		  	cur[u]=i;
		  }
		gap[d[u]]--;
		if (gap[d[u]]==0) return ans;
		d[u]=min+1;
		gap[d[u]]++;
		if (u!=s) {
			top--;
			u=edge[ss[top]^1].v;
			
	}
	}
	return ans;
}
int gcd(int a,int b){
	if (a%b==0) return b;
	return gcd(b,a%b);
}
int prime[600000]={0},numprime=0;
bool isNotPrime[1000010]={1,1};
void sushu(int N){
  long long int i;
   for (i=2;i<=N;i++) {
   	     if(! isNotPrime[i])                 
            prime[numprime ++]=i;    
        
        for(long j = 0 ; j < numprime && i * prime[j] <  N ; j ++)  
            {                 
                isNotPrime[i * prime[j]] = 1;    
            if( !(i % prime[j] ) )                   
                break;             
        }          
    }          
}
int a[10500];
char s[10];
vector<int > v[1000005];
int vis[1000005];
int main(){
	int e,t,i,j,x,y,cap,now;
	int n,m;
	sushu(1000000);
	scanf("%d",&t);
	for (e=1;e<=t;e++){
		memset(vis,0,sizeof(vis));
		for (i=0;i<numprime;i++) v[prime[i]].clear();
		init();
		scanf("%d",&n);
		for (i=1;i<=n;i++) scanf("%d",&a[i]);
		for (j=1;j<=n;j++) {
			scanf("%s",s);
		
			if (s[0]==''M'') {
				now=a[j];
				for (i=0;i<numprime&&prime[i]<=1000;i++) {
				//	printf("%d\n",now);
					if (now%prime[i]==0) v[prime[i]].push_back(j);
					while (now%prime[i]==0) now=now/prime[i];
					if (now==1||isNotPrime[now]==0) break;	
				}	
				if (isNotPrime[now]==0) v[now].push_back(j);
			} else {
				now=a[j];
				for (i=0;i<numprime&&prime[i]<=1000;i++) {
					if (now%prime[i]==0) v[prime[i]].push_back(-j);
					while (now%prime[i]==0) now=now/prime[i];
					if (now==1||isNotPrime[now]==0) break;	
				}
				if (isNotPrime[now]==0) v[now].push_back(-j);
			}
		}
		//	printf("%d %d %d\n",prime[0],v[prime[0]][0],v[prime[0]][1]);
		for (i=0;i<numprime;i++) 
		if (v[prime[i]].size()==2&&v[prime[i]][0]*v[prime[i]][1]<0) {
			
			if (v[prime[i]][0]>0) {
				
				if (vis[v[prime[i]][0]]==0) {add(0,v[prime[i]][0],1);vis[v[prime[i]][0]]=1;}
				if (vis[-v[prime[i]][1]]==0) {add(-v[prime[i]][1],n+1,1);vis[-v[prime[i]][1]]=1;}
				add(v[prime[i]][0],-v[prime[i]][1],inf);
			//	printf("%d %d\n",v[prime[i]][0],-v[prime[i]][1]);
			}
			else {
				if (vis[v[prime[i]][1]]==0) {add(0,v[prime[i]][1],1);vis[v[prime[i]][1]]=1;}
				if (vis[-v[prime[i]][0]]==0) {add(-v[prime[i]][0],n+1,1);vis[-v[prime[i]][0]]=1;}
				add(v[prime[i]][1],-v[prime[i]][0],inf);
			//	printf("%d %d\n",v[prime[i]][1],-v[prime[i]][0]);
			}
		}
		int ans=sap(0,n+1,n+2);
		printf("%d\n",n-ans);
	}
}

#include"stdio.h"
#include"string.h"
#include"vector"
#include"algorithm"
using namespace std;
vector<int> v;
int d[20];
int a[50][50];
int dp[17][2600];
int main(){
	int i,j,k,l;
	int e,t,n;
	int ans,x;
	int tmp;
	d[0]=1;
	v.clear();
	for (i=1;i<=17;i++) d[i]=d[i-1]*2;
	for (i=0;i<=d[16]-1;i++) {
		int sign=0;
		for (j=0;j<=14;j++) 
			if ((i&d[j])>0&&(i&d[j+1])>0) {sign=1;break;}
		if (sign==0) v.push_back(i);
	}
	
	l=v.size();
	scanf("%d",&t);
	for (e=1;e<=t;e++) {
		scanf("%d",&n);
		for (i=1;i<=n;i++)
		for (j=1;j<=n;j++) scanf("%d",&a[i][j]);
		memset(dp,0,sizeof(dp));
		for (i=0;i<l&&v[i]<=d[n]-1;i++) {
			for (j=1;j<=n;j++) if ((v[i]&d[j-1])>0) dp[1][i]+=a[1][j];
		//	printf("%d :%d\n",i,dp[1][v[i]]);
		}
		for (k=2;k<=n;k++) {
			for (i=0;i<l&&v[i]<=d[n]-1;i++) {
			int tmp=0;
			for (x=1;x<=n;x++) if ((v[i]&d[x-1])>0) tmp+=a[k][x];
			for (j=0;j<l&&v[j]<=d[n]-1;j++) 
			if ((v[i]&v[j])==0&&((v[i]<<1)&v[j])==0&&(v[i]&(v[j]<<1))==0) {

		    	dp[k][i]=max(dp[k][i],tmp+dp[k-1][j]);
		//    	printf("%d :%d :%d\n",k,v[i],dp[k][v[i]]);
		    }
		}
		}
		ans=0;
		for (i=0;i<l&&v[i]<=d[n]-1;i++) ans=max(dp[n][i],ans);
		printf("%d\n",ans);
	}
}

Aspect-Oriented Programming : Aspect-Oriented Programming with the RealProxy Class

A well-architected application has separate layers so different concerns don’t interact more than needed. Imagine you’re designing a loosely coupled and maintainable application,but in the middle of the development,you see some requirements that might not fit well in the architecture,such as:

• The application must have an authentication system,used before any query or update.
• The data must be validated before it’s written to the database.
• The application must have auditing and logging for sensible operations.
• The application must maintain a debugging log to check if operations are OK.
• Some operations must have their performance measured to see if they’re in the desired range.

Any of these requirements need a lot of work and,more than that,code duplication. You have to add the same code in many parts of the system,which goes against the “don’t repeat yourself” (DRY) principle and makes maintenance more difficult. Any requirement change causes a massive change in the program. When I have to add something like that in my applications,I think,“Why can’t the compiler add this repeated code in multiple places for me?” or,“I wish I had some option to ‘Add logging to this method.’”

The good news is that something like that does exist: aspect-oriented programming (AOP). It separates general code from aspects that cross the boundaries of an object or a layer. For example,the application log isn’t tied to any application layer. It applies to the whole program and should be present everywhere. That’s called a crosscutting concern.

AOP is,according to Wikipedia,“a programming paradigm that aims to increase modularity by allowing the separation of crosscutting concerns.” It deals with functionality that occurs in multiple parts of the system and separates it from the core of the application,thus improving separation of concerns while avoiding duplication of code and coupling.

In this article,I’ll explain the basics of AOP and then detail how to make it easier by using a dynamic proxy via the Microsoft .NET Framework class RealProxy.