请问高手如下代码的运算过程和结果PrivateSubCommand1Click()Dimarr1(10)AsInteger,arr2(10)AsIntegern=3Fori=1To5arr1(i)=iarr2(n)=2*n+iNextiPrintarr2(i);arr1(i)EndSub

Question

请问高手如下代码的运算过程和结果
Private Sub Command1_Click()
Dim arr1(10) As Integer,arr2(10) As Integer
n = 3
For i = 1 To 5
arr1(i) = i
arr2(n) = 2 * n + i
Next i
Print arr2(i); arr1(i)
End Sub

Answer 1

朋友你好,解释如下：
Private Sub Command1_Click()
Dim arr1(10) As Integer, arr2(10) As Integer '定义整型数组arr1,有11个元素,每个元素初始值为0；定义整型数组arr2,也是有11个元素,每个元素初始值为0
n = 3 'n是整数,赋值为3
For i = 1 To 5 'for循环,循环5次,循环了5次以后停止循环,那时i的值将为6
arr1(i) = i '给数组arr1的元素赋值,因为for循环了5次,所以依次为：arr(1)=1、arr(2)=2、arr(3)=3、arr(4)=4、arr(5)=5
arr2(n) = 2 * n + i '给数组arr2的元素赋值,因为n的值从头到尾都没有变过,所以循环5次都是给arr2(3)赋值,最后一次赋值：arr2(3)=2 * n + i=2 * 3 + 5=11,即：arr2(3)=11
Next i
Print arr2(i); arr1(i) '输出值,因为for循环结束后,i的值为6,所以输出的值是arr2(6)和arr1(6),这两个的值在程序中都没有赋值,都还是初始值0,所以程序输出0 0
End Sub