函数返回

Question

为什么此程序始终从函数year（）输出“future”的文本？

在这种情况下，B + C等于56，VAR年应归入（B + C）> 0 & &（B + C）< 1000），并返回 “罗马”，但它不是返回 “未来” ...

我得到了这一点，如果我添加此能够成功：

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

，并在我的身体机能，使条件语句检查期间。例如

if (period == "future") 

但我不明白为什么我需要这样做。我返回一个字符串，为什么我必须设置另一个变量？没有编译器错误，因此它显然不是语法的？

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

您正在将函数作为参数传递给其他函数。您需要将函数调用的结果作为参数传递给其他函数。另外，如果在字符串连接中使用int + int，则需要在括号内放置该计算。所以用（int + int）代替。

在介绍函数中，您传入的总和作为参数，但未使用它。相反，你重新计算b + c。

试试这个：

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

尝试，而不是你的if/else结构如下：

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; 
    } 
} 

你并不完全核对参数，所以它默认为 “未来” 每次。它适用于取代以前版本的此功能。