从函数VBA返回动态数组
我试图创建一个输出数组的函数。 然而,我得到了左侧的函数调用必须返回Variant或 对象如何从该函数返回动态数组从函数VBA返回动态数组,vba,Vba,我试图创建一个输出数组的函数。 然而,我得到了左侧的函数调用必须返回Variant或 对象如何从该函数返回动态数组 Public Function Fibonacci_Array(max As Integer) As Integer Dim result() As Variant ReDim result(0 To max) '' Array indices. Dim i1 As Integer Dim i2 As
Public Function Fibonacci_Array(max As Integer) As Integer
Dim result() As Variant
ReDim result(0 To max)
'' Array indices.
Dim i1 As Integer
Dim i2 As Integer
Dim i As Integer
i1 = 0
i2 = 1
'' Array values.
Dim newVal As Long
Dim prev2 As Long
Dim prev As Long
prev2 = 0
prev = 1
'' Loop through
While prev <= max
result(i1) = prev2
result(i2) = prev
newVal = prev + prev2
''Debug.Print newVal
prev2 = prev
prev = newVal
i1 = i1 + 1
i2 = i2 + 1
Wend
'' Problem here.
Fibonacci_Array() = result
End Function
公共函数Fibonacci_数组(最大为整数)为整数
Dim result()作为变量
重拨结果(0到最大值)
“”数组索引。
作为整数的Dim i1
作为整数的Dim i2
作为整数的Dim i
i1=0
i2=1
“”数组值。
暗淡的纽瓦尔像长的一样
长2倍
与上一个一样长
prev2=0
prev=1
“”循环通过
而prevVariant
是向函数传递数组或从函数传递数组时最灵活的类型
替换
Public Function Fibonacci_Array(max As Integer) As Integer
Dim result() As Variant
借
替换
Public Function Fibonacci_Array(max As Integer) As Integer
Dim result() As Variant
借
替换
Fibonacci_Array() = result
借
这将使它可以编译,但您似乎需要一些调试,因为当我输入
?Join(Fibonacci_Array(10),", ")
在即时窗口中,我得到:
0, 1, 1, 2, 3, 5, 8, , , ,
(如果您想要小于max
的斐波那契数,这可能是您想要的,但是您可能需要使用ReDim Preserve
在返回数组之前缩减数组的大小。如果您的意图是获取第一个max
斐波那契数,那么罪魁祸首是行而prevVariant
是向函数传递数组或从函数传递数组时最灵活的类型
替换
Public Function Fibonacci_Array(max As Integer) As Integer
Dim result() As Variant
借
替换
Public Function Fibonacci_Array(max As Integer) As Integer
Dim result() As Variant
借
替换
Fibonacci_Array() = result
借
这将使它可以编译,但您似乎需要一些调试,因为当我输入
?Join(Fibonacci_Array(10),", ")
在即时窗口中,我得到:
0, 1, 1, 2, 3, 5, 8, , , ,
(如果您想要小于max
的斐波那契数,这可能是您想要的,但是您可能需要使用ReDim Preserve
在返回数组之前将数组缩减为大小。如果您的意图是获取第一个max
斐波那契数,则罪魁祸首是行,而返回类型为prev应相同,并且在分配函数值时不需要括号:
Public Function Fibonacci_Array(max As Integer) As Long()
Dim result() As Long
ReDim result(0 To max)
'' Array indices.
Dim i1 As Integer
Dim i2 As Integer
Dim i As Integer
i1 = 0
i2 = 1
'' Array values.
Dim newVal As Long
Dim prev2 As Long
Dim prev As Long
prev2 = 0
prev = 1
'' Loop through
While prev <= max
result(i1) = prev2
result(i2) = prev
newVal = prev + prev2
''Debug.Print newVal
prev2 = prev
prev = newVal
i1 = i1 + 1
i2 = i2 + 1
Wend
'' Problem here.
Fibonacci_Array = result
End Function
Sub a()
Dim b() As Long
b() = Fibonacci_Array(100)
End Sub
公共函数Fibonacci_数组(最大为整数)为Long()
Dim result()的长度相同
重拨结果(0到最大值)
“”数组索引。
作为整数的Dim i1
作为整数的Dim i2
作为整数的Dim i
i1=0
i2=1
“”数组值。
暗淡的纽瓦尔像长的一样
长2倍
与上一个一样长
prev2=0
prev=1
“”循环通过
而prev您的返回类型应该相同,并且在分配函数值时不需要括号:
Public Function Fibonacci_Array(max As Integer) As Long()
Dim result() As Long
ReDim result(0 To max)
'' Array indices.
Dim i1 As Integer
Dim i2 As Integer
Dim i As Integer
i1 = 0
i2 = 1
'' Array values.
Dim newVal As Long
Dim prev2 As Long
Dim prev As Long
prev2 = 0
prev = 1
'' Loop through
While prev <= max
result(i1) = prev2
result(i2) = prev
newVal = prev + prev2
''Debug.Print newVal
prev2 = prev
prev = newVal
i1 = i1 + 1
i2 = i2 + 1
Wend
'' Problem here.
Fibonacci_Array = result
End Function
Sub a()
Dim b() As Long
b() = Fibonacci_Array(100)
End Sub
公共函数Fibonacci_数组(最大为整数)为Long()
Dim result()的长度相同
重拨结果(0到最大值)
“”数组索引。
作为整数的Dim i1
作为整数的Dim i2
作为整数的Dim i
i1=0
i2=1
“”数组值。
暗淡的纽瓦尔像长的一样
长2倍
与上一个一样长
prev2=0
prev=1
“”循环通过
上一篇我希望我能给这个回答3票以上!一票正确回答我的问题,一票给出输出斐波那契序列的替代方法,一票提到比奈公式。我希望我能给这个回答3票以上!一票正确回答我的问题,一票给出输出斐波那契序列的替代方法ce,还有一个提到比奈公式。