delegate的衍化过程 分类： .NET 2014-05-14 17:03 301人阅读 评论(0) 收藏
class Program
{
delegate bool Foo(int n);
static bool foo1(int n)
{
return true;
}
static bool foo2(int n)
{
return n % 2 == 0 ? true : false;
}
static bool foo3(int n)
{
return n % 2 != 0 ? true : false;
}
static void Main(string[] args)
{
Console.WriteLine(Sum(10, foo1));//10以内数相加
Console.WriteLine(Sum(10, foo2));//10以内偶数相加
Console.WriteLine(Sum(10, foo3));//10以内奇数相加
Console.Read();
}
static int Sum(int n, Foo foo)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
if (foo(i))
{
sum += i;
}
}
return sum;
}
}
class Program
{
delegate bool Foo(int n);
static void Main(string[] args)
{
Console.WriteLine(Sum(10, delegate(int a) { return true; })); //10以内数相加
Console.WriteLine(Sum(10, delegate(int a) { return a % 2 == 0; }));//10以内偶数相加
Console.WriteLine(Sum(10, delegate(int a) { return a % 2 != 0; }));//10以内奇数相加
Console.Read();
}
static int Sum(int n, Foo foo)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
if (foo(i))
{
sum += i;
}
}
return sum;
}
}
class Program
{
delegate bool Foo(int n);
static void Main(string[] args)
{
//进化为更简洁的Lambda表达式来处理
Console.WriteLine(Sum(10, (i) => true)); //10以内数相加
Console.WriteLine(Sum(10, (i) => i % 2 == 0));//10以内偶数相加
Console.WriteLine(Sum(10, (i) => i % 2 != 0));//10以内奇数相加
Console.Read();
}
static int Sum(int n, Foo foo)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
if (foo(i))
{
sum += i;
}
}
return sum;
}
}
