函數返回

Question

爲什麼此程序始終從函數year（）輸出「future」的文本？

在這種情況下，B + C等於56，VAR年應歸入（B + C）> 0 & &（B + C）< 1000），並返回 「羅馬」，但它不是返回 「未來」 ...

我得到了這一點，如果我添加此能夠成功：

var period:String = (year(b,c)); 

，並在我的身體機能，使條件語句檢查期間。例如

if (period == "future") 

但我不明白爲什麼我需要這樣做。我返回一個字符串，爲什麼我必須設置另一個變量？沒有編譯器錯誤，因此它顯然不是語法的？

var a:String = "Tim"; 
var b:int = 50; //CHANGE TO ANY INT YOU WANT 
var c:int = 6; //CHANGE TO ANY INT YOU WANT 
var d:String = "Kyle"; 
var sum:int = b+c; 

function friend(d:String, a:String):String 
{ 
    return d+" and "+a; 
} 

function year(b:int, c:int):String 
{ 
    if((b+c) > 2000) 
     return "future"; 
    else if((b+c)> 1000 && b+c< 2000) 
     return "colonial"; 
    else if((b+c) > 0 && (b+c) < 1000) 
     return "roman"; 
    else if((b+c) < 0) 
     return "medieval"; 
    else 
     return "fail"; 

} 


function intro(sum, friend):String 
{ 
    return "Once upon a time, in the year "+ b+c +", "+friend; 
} 

function body(year):String 
{ 
    if ("future") 
     return " saw a flying saucer and descided they wanted do be an alien."; 
    else if ("colonial") 
     return " just got off the the Mayflower and descided they wanted to eat some turkey."; 
    else if ("roman") 
     return " are taking a break after a fierce battle with the Romans."; 
    else if ("medieval") 
     return " saved the princess in shining armor after slaying the dragon."; 
    else if ("fail") 
     return " just got an F on their exam."; 
    else 
     return " just got an F on their test.";    
} 
trace (b+c); 
trace(intro(sum, friend(d, a)) + body(year)); 

Answer 1

您正在將函數作爲參數傳遞給其他函數。您需要將函數調用的結果作爲參數傳遞給其他函數。另外，如果在字符串連接中使用int + int，則需要在括號內放置該計算。所以用（int + int）代替。

在介紹函數中，您傳入的總和作爲參數，但未使用它。相反，你重新計算b + c。

試試這個：

var a:String = "Tim"; 
var b:int = 50; //CHANGE TO ANY INT YOU WANT 
var c:int = 6; //CHANGE TO ANY INT YOU WANT 
var d:String = "Kyle"; 
var sum:int = b+c; 

function friend(d:String, a:String):String 
{ 
    return d+" and "+a; 
} 

function year(b:int, c:int):String 
{ 
    if((b+c) > 2000) 
     return "future"; 
    else if((b+c)> 1000 && b+c< 2000) 
     return "colonial"; 
    else if((b+c) > 0 && (b+c) < 1000) 
     return "roman"; 
    else if((b+c) < 0) 
     return "medieval"; 
    else 
     return "fail"; 

} 


function intro(sum:int, friend:String):String 
{ 
    return "Once upon a time, in the year "+ sum +", "+friend; 
} 

function body(year:String):String 
{ 
    if ("future") 
     return " saw a flying saucer and descided they wanted do be an alien."; 
    else if ("colonial") 
     return " just got off the the Mayflower and descided they wanted to eat some turkey."; 
    else if ("roman") 
     return " are taking a break after a fierce battle with the Romans."; 
    else if ("medieval") 
     return " saved the princess in shining armor after slaying the dragon."; 
    else if ("fail") 
     return " just got an F on their exam."; 
    else 
     return " just got an F on their test.";    
} 
trace (b+c); 
trace(intro(sum, friend(d, a)) + body(year(b, c))); 

Answer 2

嘗試，而不是你的if/else結構如下：

function body(year:String):String 
{ 
    switch(year) 
    { 
     case "future": 
     return " saw a flying saucer and descided they wanted do be an alien."; 
     break; 
     case "colonial": 
     return " just got off the the Mayflower and descided they wanted to eat some turkey."; 
     break; 
     case "roman": 
     return " are taking a break after a fierce battle with the Romans."; 
     break; 
     case "medieval": 
     return " saved the princess in shining armor after slaying the dragon."; 
     break; 
     case "fail": 
     return " just got an F on their exam."; 
     break; 
     default: 
     return " just got an F on their test."; 
     break; 
    } 
} 

你並不完全覈對參數，所以它默認爲 「未來」 每次。它適用於取代以前版本的此功能。