# Func<T1,T2,T3,T4,T5,T6,T7,T8,TResult> 委托

## 定义

``````generic <typename T1, typename T2, typename T3, typename T4, typename T5, typename T6, typename T7, typename T8, typename TResult>
public delegate TResult Func(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5, T6 arg6, T7 arg7, T8 arg8);``````
``public delegate TResult Func<in T1,in T2,in T3,in T4,in T5,in T6,in T7,in T8,out TResult>(T1 arg1, T2 arg2, T3 arg3, T4 arg4, T5 arg5, T6 arg6, T7 arg7, T8 arg8);``
``type Func<'T1, 'T2, 'T3, 'T4, 'T5, 'T6, 'T7, 'T8, 'Result> = delegate of 'T1 * 'T2 * 'T3 * 'T4 * 'T5 * 'T6 * 'T7 * 'T8 -> 'Result``
``Public Delegate Function Func(Of In T1, In T2, In T3, In T4, In T5, In T6, In T7, In T8, Out TResult)(arg1 As T1, arg2 As T2, arg3 As T3, arg4 As T4, arg5 As T5, arg6 As T6, arg7 As T7, arg8 As T8) As TResult ``

T1

T2

T3

T4

T5

T6

T7

T8

TResult

arg1
T1

arg2
T2

arg3
T3

arg4
T4

arg5
T5

arg6
T6

arg7
T7

arg8
T8

#### 返回值

TResult

Func<T1,T2,T3,T4,T5,T6,T7,T8,TResult>

## 注解

Lambda 表达式的基础类型是一个泛型 `Func` 委托。The underlying type of a lambda expression is one of the generic `Func` delegates. 这样，便可以将 lambda 表达式作为参数传递，而无需将其显式分配给委托。This makes it possible to pass a lambda expression as a parameter without explicitly assigning it to a delegate.

## 扩展方法

 获取指示指定委托表示的方法的对象。Gets an object that represents the method represented by the specified delegate.