了,每次返回一个新的值 : public static T If<T>(this T t, Predicate<T> predicate, Func<T, T> func) where T : struct
{
return predicate(t) ? func(t) : t;
}
调用代码也要修改:
public static void Test2()
{
//扩展方式
int int0 = -121;
int int1 = int0.If(i => i < 0, i => -i)
.If(i => i > 100, i => i - 100)
.If(i => i % 2 == 1, i => i - 1);
//常规方式
int int3 = -121;
if (int3 < 0) int3 = -int3;
if (int3 > 100) int3 -= 100;
if (int3 % 2 == 1) int3--;
}
引用类型及值类型的扩展我们已经完成,用string来测试一下吧,如下:
public static void Test3()
{
//从邮箱变换成主页
string email = "ldp615@163.com";
string page = email.If(s => s.Contains("@"), s => s.Substring(0, s.IndexOf("@")))
.If(s =>! s.StartsWith("www."), s => s = "www." + s)
.If(s =>! s.EndsWith(".com"), s => s += ".com");
}
但编译不通过,会提示错误:

这个错误比较怪,我们写了两个扩展,一个是给值类型的,一个给引用类型,可string类型在这里都不行。这个原因我说不清楚了,还留给园子里高手们吧。
不过专门为string写个扩展,这个问题可以化解,如下:
public static string If(this string s, Predicate<string> predicate, Func<string, string> func)
{
return predicate(s) ? func(s) : s;
}
看来扩展方法也是有优先级的:对同一个类进行多次扩展,扩展方法相名,参数也等效(数量、顺序相同),非泛版扩展要比泛型版扩展优先级高。
下面再来看一段swith代码,很啰嗦的!这里是为了引出扩展硬写出来的:
public static void Test4()
{
string englishName = "apple";
string chineseName = string.Empty;
switch (englishName)
{
case "apple":
chineseName = "苹果";
return;
case "orange":
chineseName = "桔子";
return;
case "banana":
chineseName = "香蕉";
return;
case "pear":
chineseName = "梨";
break;
default:
chineseName = "未知";
break;
}
Console.WriteLine(chineseName);
}
我们把这种方式用扩展方法来完成:
public static TOutput Switch<TOutput, TInput>(this TInput input, IEnumerable<TInput> inputSource, IEnumerable<TOutput> outputSource, TOutput defaultOutput)
{
IEnumerator<TInput> inputIterator = inputSource.GetEnumerator();
IEnumerator<TOutput> outputIterator = outputSource.GetEnumerator();
TOutput result = defaultOutput;
while (inputIterator.MoveNext())
{
if (outputIterator.MoveNext())
{
if (input.Equals(inputIterator.Current))
{
result = outputIterator.Current;
break;
}
}
else break;
}
return result;
}
下面的Test5是调用:
public static void Test5()
{
string englishName = "apple";
string chineseName = englishName.Switch(
new string[] { "apple", "orange", "banana", "pear" },
new string[] { "苹果", "桔子", "香蕉", "梨" },
"未知"
);
Console.WriteLine(chineseName);
}
简单清晰明了!
最后是一个对while的扩展封装:
public static void While<T>(this T t, Predicate<T> predicate, Action<T> action) where T: class
{
while (predicate(t)) action(t);
}
调用代码:
public static void Test6()
{
People people = new People { Name = "Wretch" };
people.While(
p => p.WorkCount < 7,
p => p.Work()
);
people.Rest();
}
这里又“召唤”了一个人,不让吃喝不让休息,连续工作7次...
这while扩展中只能执行一个Action,不太好,我们用params改进一下:
public static void While<T>(this T t, Predicate<T> predicate, params Action<T>[] actions) where T : class
{
while (predicate(t))
{
foreach (var action in actions)
action(t);
}
}
再来调用,可以在循环中执行多个操作了,这次舒服工作完吃饭喝水休息,再来工作...
public static void Test7()
{
People people = new People { Name = "Wretch" };
people.While(
p => p.WorkCount < 7,
p => p.Work(),
p => p.Eat(),
p => p.Drink(),
p => p.Rest()
);
people.Rest();
}
当然前面的If也可以这样的,这里只写出一个:

public static T If<T>(this T t, Predicate<T> predicate, params Action<T>[] actions) where T : class
{
if (t == null) throw new ArgumentNullException();
if (predicate(t))
{
foreach (var action in actions)
action(t);
}
return t;
}