Tuple<T1,T2,T3,T4,T5>(T1, T2, T3, T4, T5) 构造函数

定义

初始化 Tuple<T1,T2,T3,T4,T5> 类的新实例。

public:
 Tuple(T1 item1, T2 item2, T3 item3, T4 item4, T5 item5);
public Tuple (T1 item1, T2 item2, T3 item3, T4 item4, T5 item5);
new Tuple<'T1, 'T2, 'T3, 'T4, 'T5> : 'T1 * 'T2 * 'T3 * 'T4 * 'T5 -> Tuple<'T1, 'T2, 'T3, 'T4, 'T5>
Public Sub New (item1 As T1, item2 As T2, item3 As T3, item4 As T4, item5 As T5)

参数

item1
T1

此元组的第一个组件的值。

item2
T2

此元组的第二个组件的值。

item3
T3

此元组的第三个组件的值。

item4
T4

元组的第四个分量的值。

item5
T5

元组的第五个分量的值。

注解

还可以使用静态 Tuple.Create<T1,T2,T3,T4,T5>(T1, T2, T3, T4, T5) 方法实例化 5 元组对象,而无需显式指定其组件的类型。 以下示例使用 Tuple.Create<T1,T2,T3,T4,T5>(T1, T2, T3, T4, T5) 方法实例化一个 5 元组,其第一个组件为 类型 String ,其其余四个组件的类型 Int32为 。

var tuple5 = Tuple.Create("New York", 1990, 7322564, 2000, 8008278);
Console.WriteLine("{0}: {1:N0} in {2}, {3:N0} in {4}",
                  tuple5.Item1, tuple5.Item3, tuple5.Item2,
                  tuple5.Item5, tuple5.Item4);
// Displays New York: 7,322,564 in 1990, 8,008,278 in 2000
let tuple5 =
    Tuple.Create("New York", 1990, 7322564, 2000, 8008278)

printfn $"{tuple5.Item1}: {tuple5.Item3:N0} in {tuple5.Item2}, {tuple5.Item5:N0} in {tuple5.Item4}"
// Displays New York: 7,322,564 in 1990, 8,008,278 in 2000
Dim tuple5 = Tuple.Create("New York", 1990, 7322564, 2000, 
                          8008278)
Console.WriteLine("{0}: {1:N0} in {2}, {3:N0} in {4}",
                  tuple5.Item1, tuple5.Item3, tuple5.Item2,
                  tuple5.Item5, tuple5.Item4)
' Displays New York: 7,322,564 in 1990, 8,008,278 in 2000

这等效于对类构造函数的 Tuple<T1,T2,T3,T4,T5> 以下调用。

var tuple5 = new Tuple<string, int, int, int, int>
                      ("New York", 1990, 7322564, 2000, 8008278);
Console.WriteLine("{0}: {1:N0} in {2}, {3:N0} in {4}",
                  tuple5.Item1, tuple5.Item3, tuple5.Item2,
                  tuple5.Item5, tuple5.Item4);
// Displays New York: 7,322,564 in 1990, 8,008,278 in 2000
let tuple5 =
    Tuple<string, int, int, int, int>("New York", 1990, 7322564, 2000, 8008278)

printfn $"{tuple5.Item1}: {tuple5.Item3:N0} in {tuple5.Item2}, {tuple5.Item5:N0} in {tuple5.Item4}"
// Displays New York: 7,322,564 in 1990, 8,008,278 in 2000
Dim tuple5 = New Tuple(Of String, Integer, Integer, 
                       Integer, Integer) _
                       ("New York", 1990, 7322564, 2000, 8008278)
Console.WriteLine("{0}: {1:N0} in {2}, {3:N0} in {4}",
                  tuple5.Item1, tuple5.Item3, tuple5.Item2,
                  tuple5.Item5, tuple5.Item4)
' Displays New York: 7,322,564 in 1990, 8,008,278 in 2000

适用于