共用方式為


Tuple<T1,T2,T3,T4,T5>(T1, T2, T3, T4, T5) 建構函式

定義

初始化 Tuple<T1,T2,T3,T4,T5> 類別的新執行個體。

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)

參數

item1
T1

Tuple 第 1 個元件的值。

item2
T2

Tuple 第 2 個元件的值。

item3
T3

Tuple 第 3 個元件的值。

item4
T4

元組第四個元件的值。

item5
T5

Tuple 第 5 個元件的值。

備註

您也可以使用靜態 Tuple.Create<T1,T2,T3,T4,T5>(T1, T2, T3, T4, T5) 方法來具現化 5 元組物件,而不需要明確指定其元件的類型。 下列範例會 Tuple.Create<T1,T2,T3,T4,T5>(T1, T2, T3, T4, T5) 使用 方法來具現化第一個元件為 類型的 String 5 元組,而其餘四個元件的類型為 Int32

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

這相當於對類別建構函式的 Tuple<T1,T2,T3,T4,T5> 下列呼叫。

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

適用於