VB 十进制转二进制 代码问题 急！！

Question

Private Sub Command1_Click()
Dim n As Integer, num As Long
Dim ch As String, Bin() As String, r As Integer, i As Integer
num = Val(Text1)
n = 0
Do
r = num Mod 2
n = n + 1
ReDim Bin(n)
Loop Until r = 0
For i = UBound(Bin) To 1 Step -1
ch = ch & Bin(i)
Next i
Text2 = ch
End Sub

哪里出了错？跪求高手帮忙修改!

Answer 1

这样：

Private Sub Command1_Click()

 Dim n As Integer, num As Long

 Dim ch As String, Bin() As String, r As Integer, i As Integer

 num = Val(Text1.Text)

 n = 0

 r = num

 Do

   n = n + 1

   ReDim Preserve Bin(n)

   Bin(n) = r Mod 2

   r = r \ 2

 Loop Until r < 1

 For i = UBound(Bin) To 1 Step -1

   ch = ch & Bin(i)

 Next i

 Text2.Text = ch

End Sub

redim数组变量没有给其赋值。

扩展资料：

注意事项

一个十进制转二进制的函数：

Public Function D2B(D As Integer) As String

    D2B = ""

    Do While D > 0

        D2B = D Mod 2 & D2B

        D = D \ 2

    Loop

End Function 

Private Sub Form_Load()

    Me.Caption = D2B(8)

End Sub

另外的办法：

Function dToTwo(n As Integer) As String

Dim S As String

S = ""

Do While n <> 0 

    a = n Mod 2 

    n = n \ 2 

    S = Chr(48 + a) & S

Loop

dToTwo= S

End Function

Answer 2

Private Sub Command1_Click()
r = Val(Text1)
n = 0
Do
ch = r Mod 2 & ch
r = r \ 2
Loop Until r = 0
Text2 = ch
End Sub

你redim 数组变量 却不给其赋值 不是白搭么
再者你若是要 动态REDIM 也要加Preserve 这样才可保留住之前的数据
redim Preserve bin(n)

再者 r = num Mod 2 永远都是初始值NUM 去求余 那R值不都永远一样么
要是Num是奇数 不就死循环了

Answer 3

首先你要明白十进制转二进制的方法
如 100 转成2进制
用辗转相除法
100
除2 50 余数0
除2 25 余数0
除2 12 余数1
除2 6 余数0
除2 3 余数0
除2 1 余数1
除2 0 余数1
把相除过程中产生的余数倒序看 1100100 这就是100变成2进制的值
再看程序
Public Function DECtoBIN(Dec As Long) As String
DECtoBIN = ""
Do While Dec > 0 //这个循环就是辗转相除法的循环
DECtoBIN = Dec Mod 2 & DECtoBIN //DECtoBIN 是每次产生的余数连成的字符串
Dec = Dec \ 2 //dec为下次要进行除法的被除数。
Loop //注意这里字符串连接是倒序连接
End Function
同理看下面程序
Dim a%(10),m%,i%
M=val(inputbox(""))
i=0
do while m<>0 //这个也是辗转相除法的循环
a(i)=m mod 2 //用数组保留除得的余数
m=m\2 //m为下次要进行除法的被除数
i=i+1
loop
for k=i-1 to 0 step-1 //倒序输出余数
print a(k);
next追问

额，我不要其他方法。。。就想知道自己的方法哪里出错了

追答

Private Sub Command1_Click()
Dim n As Integer, num As Long
Dim ch As String, Bin() As String, r As Integer, i As Integer
num = Val(Text1.Text)
n = 0
r = num
Do
n = n + 1
ReDim Preserve Bin(n)
Bin(n) = r Mod 2
r = r \ 2
Loop Until r < 1
For i = UBound(Bin) To 1 Step -1
ch = ch & Bin(i)
Next i
Text2.Text = ch
End Sub

本回答被提问者采纳

Answer 4

n=12345
do while(n>2)
s=s & (n and 1)
n=n\2
loop
s=strreverse(s)
msgbox s