Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load

    End Sub
    Sub procmin(ByRef a%(), ByVal min%)
        Dim i%
        min = a(0)
        For i = 0 To UBound(a)
            If a(i) < min Then min = a(i)

        Next
    End Sub
    Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
        Dim a(9) As Integer, j As Integer, min As Integer
        For j = 0 To UBound(a)
            a(j) = Int(Rnd() * 100 + (-400))
            Call procmin(a(j), min)
        Next
        Label1.Text = "min=" & min

    End Sub
End Class

错误之处：
一、Call procmin(a(j), min)这句有错。从上面定义的过程procmin可以看出，传入的第一个参数是数组名或认为是指针，而你实际调用的时候传入的是a(j),这是一个整数，类型不匹配
二、还在这一句，如果你要调用上面定义好的过程来判断数组a里的最小值，不需要在给数组a赋值的For循环中调用此过程，而应在For循环结束后调用此过程
三、procmin这是过程没有返回值，所以要把第二个参数min的传值类型改为byref,这样在程序的最后才能用min的值正确显示数组中的最小值
改后的代码如下：

Public Class Form1
    Sub procmin(ByRef a%(), ByRef min%)
        Dim i%
        min = a(0)
        For i = 0 To UBound(a)
            If a(i) < min Then min = a(i)

        Next
    End Sub
    Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
        Dim a(9) As Integer, j As Integer, min As Integer
        For j = 0 To UBound(a)
            a(j) = Int(Rnd() * 100 + (-400))


        Next
        MsgBox(a(0) & " " & a(1) & " " & a(2) & " " & a(3) & " " & a(4) & " " & a(5) & " " & a(6) & " " & a(7) & " " & a(8) & " " & a(9) & " ")
        Call procmin(a, min)
        Label1.Text = "min=" & min

    End Sub

End Class

msgbox函数可以查看取得的随机数，以验证程序的正确性。



[ 本帖最后由 wei855198 于 2010-4-27 21:48 编辑 ]

大哥嘿嘿 谢谢  嘿嘿  我后来懂了 所以把分取消了 我再发一个 你来回几个字 我一定要把分给你

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