vb.net 函數(shù)或者過程作為數(shù)組

thr_test(i) = New System.Threading.Thread(New System.Threading.ParameterizedThreadStart(Sub()

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直接寫過程代碼

End Sub))

vb.net 返回一個數(shù)組的函數(shù)

public function createstringarr() as string()

return new string(){"d1","d2","d3","d4"}

end function

VB.net 如何編寫一個可以返回?cái)?shù)組的函數(shù)（過程）？

public function createstringarr() as string()

return new string(){"d1","d2","d3","d4"}

end function

vb.net 編寫一個函數(shù)

注意：參數(shù)為動態(tài)數(shù)組；

Private Function MyF(ByRef d() As Integer)

ReDim d(4, 13) As Integer

Dim i As Integer

Dim j As Integer

Dim n As Integer

Dim MyNum(4) As Integer

For i = 1 To 4

MyNum(i) = 0

Next i

Randomize

For i = 1 To 4

For j = 1 To 13

n = Int(Rnd * 4 + 1)

Do While MyNum(n) = 13

n = Int(Rnd * 4 + 1)

Loop

d(i, j) = n

MyNum(n) = MyNum(n) + 1

Next j

Next i

End Function

vb.net中返回一個數(shù)組的函數(shù)的問題

給你一個簡單的示例：

Option?Explicit

Private?Function?GetRndInt()?As?Integer()

'?返回一個?Integer類型的數(shù)組

Dim?i?As?Long,?aTemp()?As?Integer

Randomize

ReDim?aTemp(15)

For?i?=?0?To?15

aTemp(i)?=?Rnd()?*?500

Next

GetRndInt?=?aTemp

End?Function

Private?Sub?Command1_Click()

Dim?aTemp()?As?Integer

Dim?i?As?Long

Me.Cls

aTemp?=?GetRndInt

Print?"隨機(jī)產(chǎn)生的數(shù)據(jù)為："

For?i?=?0?To?UBound(aTemp)

Print?i?+?1,?aTemp(i)

Next

End?Sub

運(yùn)行效果：