vba调JAVA并接收返回值_如何从VBA函数返回结果
慕萊塢7535251
只是將返回值設置為函數名仍然與Java(或其他)語句不完全相同return,因為在java中,return退出函數,如下所示:public?int?test(int?x)?{
????if?(x?==?1)?{
????????return?1;?//?exits?immediately
????}
????//?still?here??return?0?as?default.
????return?0;}在VB中,如果未在函數末尾設置返回值,則精確等效項需要兩行。因此,在VB中,確切的推論看起來像這樣:Public?Function?test(ByVal?x?As?Integer)?As?Integer
????If?x?=?1?Then
????????test?=?1?'?does?not?exit?immediately.?You?must?manually?terminate...
????????Exit?Function?'?to?exit
????End?If
????'?Still?here??return?0?as?default.
????test?=?0
????'?no?need?for?an?Exit?Function?because?we're?about?to?exit?anyway.End?Function既然如此,那么知道你可以像使用方法中的任何其他變量一樣使用return變量也是很好的。像這樣:Public?Function?test(ByVal?x?As?Integer)?As?Integer
????test?=?x?'?
????If?test?<>?1?Then?'?Test?the?currently?set?return?value
????????test?=?0?'?Reset?the?return?value?to?a?*new*?value
????End?IfEnd?Function或者,返回變量如何工作的極端例子(但不一定是你應該如何實際編碼的一個很好的例子) - 那個會讓你夜不能寐的一個例子:Public?Function?test(ByVal?x?As?Integer)?As?Integer
????test?=?x?'?
????If?test?>?0?Then
????????'?RECURSIVE?CALL...WITH?THE?RETURN?VALUE?AS?AN?ARGUMENT,
????????'?AND?THE?RESULT?RESETTING?THE?RETURN?VALUE.
????????test?=?test(test?-?1)
????End?IfEnd?Function
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