比较菜的问题，但我有难度!!

[fuke]

dim str as string
dim code(2)as string
dim index_code as long
　code(0)="+"
code(1)="-"
code(2)="*"
for index_code=0 to 2 step 1
str="3" & code &"4"
....
next index_code
问题：请问如何得到"3" & code & "4"的数字值

[jaek999]

dim str as string
dim code(2)as string
dim index_code as long
　code(0)="+"
code(1)="-"
code(2)="*"
for index_code= 0 to 2
select case code(index_code)
case "+"
str=cstr$(3+4)
case "-"
str=cstr$(3-4)
case "*"
str=cstr$(3*4)
end select
next index_code

[lihonggen0]

Dim str As String
Dim num As Double
Dim code(2) As String
Dim index_code As Long
code(0) = "+"
code(1) = "-"
code(2) = "*"
For index_code = 0 To 2 Step 1
Select Case code(0)
Case "+"
num = Val("3") + Val("4")
Case "-"
num = Val("3") - Val("4")
Case "*"
num = Val("3") * Val("4")
End Select
Next index_code

[lihonggen0]

Dim str As String
Dim num As Double
Dim code(2) As String
Dim index_code As Long
code(0) = "+"
code(1) = "-"
code(2) = "*"
For index_code = 0 To 2 Step 1
Select Case code(0)
Case "+"
num = Val("3") + Val("4")
Case "-"
num = Val("3") - Val("4")
Case "*"
num = Val("3") * Val("4")
End Select
Next index_code

[jianglifen]

取出str的值，不过我觉得你用字符型str的值变成 34了，而不是3+4=7的值

dim str as interger
str=val(3 code(index_code) 4)

[CloneCenter]

str=val(3 code(index_code) 4)
这样也可以？？？

怎么感觉楼上似乎是个 MM？

[stiffe01]

本身用str作变量就错了！

[sikeen]

dim str as string
dim code(2)as string
dim index_code as long
code(0)="+"
code(1)="-"
code(2)="*"
str = "0"

for index_code=0 to 2 step 1
select case code(index_code)
case "+"
str = cstr(cdbl(str) + 3 + 4)
case "-"
str = cstr(cdbl(str) + 3 - 4)
case "*"
str = cstr(cdbl(str) + 3 * 4)
end select

....
next index_code

[CityhunterID]

同意楼上的说

[IwantFlay]

for index_code=0 to 2 step 1
str=str & code(index_code)
next index_code

str="3" & str & "4"