我写了一个函数
Public Function dzh(du As Double) As Double()
Dim rst(3) As Double
Dim tem As Double
rst(0) = Fix(du)
tem = (du - rst(0)) * 60
rst(1) = Fix(tem)
rst(2) = (tem - rst(1)) * 60
dzh = rst
End Function
调用函数时参数值为105.26,为什么du-rst(0)等于0.260000000005而不等于0.26呢?怎么解决啊?
Public Function dzh(du As Double) As Double()
Dim rst(3) As Double
Dim tem As Double
rst(0) = Fix(du)
tem = (du - rst(0)) * 60
rst(1) = Fix(tem)
rst(2) = (tem - rst(1)) * 60
dzh = rst
End Function
调用函数时参数值为105.26,为什么du-rst(0)等于0.260000000005而不等于0.26呢?怎么解决啊?
解决方案 »
免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货