在for循环中使用switch语句

Question

我有一个关于weather的问题，它在for循环中使用switch语句是合法的。它写在Eclipse下面的方式给了我一个错误，不会让我在for循环的第二条语句中使用switch（n）。我想要做的是写...

这位老人，他打了1
他在我的拇指上打了个小诀窍。

用for循环和switch语句解决以下问题有更好的方法吗？我想写出不同的数字和不同的情况下的段落10。所以1个拇指，2鞋，3膝盖....直到代替switch语句10.

import acm.program.*; 

public class SingSong extends ConsoleProgram { 

    public void run() { 
     for (int n = 1; n <= 10; n++) { 
      println("This old man, he played " + n); 
      println("He played knick-knack on my" + switch(n)); 
      println("With a knick-knack, paddy-whack,"); 
      println("Give your dog a bone"); 
      println("This old man came rolling home"); 

      switch (n) {    
       case 1 : println("thumb"); break; 
       case 2 : println("shoe"); break; 
       case 3 : println("knee"); break; 
       case 4 : println("door"); break; 
       case 5 : println("hive"); break; 
       case 6 : println("sticks"); break; 
       case 7 : println("heaven"); break; 
       case 8 : println ("pate"); break; 
       case 9 : println("spine"); break; 
       case 10 : println("shin"); break; 
      } 
     } 
    } 
} 

Answer 1

你想要的是一个方法，你会打电话：

private String which(int n) { 
    switch (n) { 
    case 1 : return "thumb"; 
    case 2 : return "show"; 
    case 3 : return "knee"; 
    case 4 : return "door"; 
    case 5 : return "hive"; 
    case 6 : return "sticks"; 
    case 7 : return "heaven"; 
    case 8 : return "pate"; 
    case 9 : return "spine"; 
    case 10 : return "shin"; 
    default: return ""; 

    } 
} 

然后用

println("He played knick-knack on my" + which(n)); 

另一个替换

println("He played knick-knack on my" + switch(n)); 

，更succint的方法是创建一个字符串数组：

static final String[] which = { 
    "thumb","show","knee","door","hive", "sticks","heaven","pate","spine","shin"}; 

，然后只用

println("He played knick-knack on my" + which[n]); 

Answer 2

使用if-else。

if(n==1){ 
... 
}else if(n==2){ 
... 
} 

Answer 3

println("He played knick-knack on my" + switch(n)); 

这难道不使你的交换机的呼叫。开关不是一种方法。 为了得到所需的输出，

println("This old man, he played " + n); 
      println("He played knick-knack on my"); 

    switch (n){ 

      case 1 : print("thumb"); break; 
      case 2 : println("shoe"); break; 
      case 3 : println("knee"); break; 
      case 4 : println("door"); break; 
      case 5 : println("hive"); break; 
      case 6 : println("sticks"); break; 
      case 7 : println("heaven"); break; 
      case 8 : println ("pate"); break; 
      case 9 : println("spine"); break; 
      case 10 : println("shin"); break; 

      } 

甚至更​​好地遵循@Marko Topolnik办法，我会说是最好的办法。

Answer 4

for循环中的switch/case语句没问题，问题在于switch不是函数调用。尝试是这样的：

string obj; 
    switch (n){ 

    case 1 : obj="thumb"; break; 
    case 2 : obj="show"; break; 
    //Etc 
    } 


    } 
    println("This old man, he played " + n); 
    println("He played knick-knack on my" + obj); 
    println("With a knick-knack, paddy-whack,"); 
    println("Give your dog a bone"); 
    println("This old man came roiling home"); 

Answer 5

创建Map<Integer, String>方含整串映射和使用Map.get()获得字符串：

Map<Integer, String> m = new HashMap<Integer, String>(); 
m.put(1, "thumb"); 
// etc 
m.put(10, "shin"); 

println("He played knick-knack on my" + m.get(n)); 

Answer 6

你不能做到这一点：

println("He played knick-knack on my" + switch(n)); 

如果你想让你的代码在很大程度上相同，你应该委托给一个执行切换的方法。

替代解决方案将包括一系列if/else if（有点冗长），或者可能是充当查找表的数组。

例如

// note arrays are indexed from 0 
String[] arrayOfObjects = new String[]{"thumb", ....}; 
println("He played knick-knack on my" + arrayOfObjects(n-1)); 

我想也许Map是矫枉过正在这种情况下，除非你有不连续的分度（不太可能）

Answer 7

开关（N）不是一个函数，不能这样使用。对编译器来说，这就像启动一个switch语句，但是完全错误，因此它返回一个错误。我认为要正确地做到这一点，你应该把案例切换到另一种方法，或者直接放在第二行之后。像这样：

import acm.program.*; 

public class SingSong extends ConsoleProgram { 

public void run(){ 
for (int n = 1; n <= 10; n++){ 

    println("This old man, he played " + n); 
    print("He played knick-knack on my "); 
    switch (n){ 
    case 1 : print("thumb"); break; 
    case 2 : print("show"); break; 
    case 3 : print("knee"); break; 
    case 4 : print("door"); break; 
    case 5 : print("hive"); break; 
    case 6 : print("sticks"); break; 
    case 7 : print("heaven"); break; 
    case 8 : print("pate"); break; 
    case 9 : print("spine"); break; 
    case 10 : print("shin"); break; 
    } 
    println("With a knick-knack, paddy-whack,"); 
    println("Give your dog a bone"); 
    println("This old man came roiling home"); 

     } 
     } 

     } 

这应该有效。或者用不同的方法：

import acm.program.*; 


public class SingSong extends ConsoleProgram { 
public void run(){ 

for (int n = 1; n <= 10; n++){ 

    println("This old man, he played " + n); 
    print("He played knick-knack on my"); 
    switcher(n); 
    print("\n"); 
    println("With a knick-knack, paddy-whack,"); 
    println("Give your dog a bone"); 
    println("This old man came roiling home"); 
} 
} 

    private void switcher(int n){ 
    switch(n){ 
    case 1 : print("thumb"); break; 
    case 2 : print("show"); break; 
    case 3 : print("knee"); break; 
    case 4 : print("door"); break; 
    case 5 : print("hive"); break; 
    case 6 : print("sticks"); break; 
    case 7 : print("heaven"); break; 
    case 8 : print ("pate"); break; 
    case 9 : print("spine"); break; 
    case 10 : print("shin"); break; 


     } 
     } 

     } 

Answer 8

您正在使用switch之类的方法。做下列变化，它会工作：

import acm.program.*; 

public class SingSong extends ConsoleProgram 
{ 
    public void run() 
    { 
     for (int n = 1; n <= 10; n++) 
     { 
      println("This old man, he played " + n); 
      println("He played knick-knack on my" + getValue(n)); 
      println("With a knick-knack, paddy-whack,"); 
      println("Give your dog a bone"); 
      println("This old man came roiling home"); 
     } 
    } 

    private String getValue(int n) 
    { 
     switch (n) 
     { 
      case 1 : return "thumb"; break; 
      case 2 : return "shoe"; break; 
      case 3 : return "knee"; break; 
      case 4 : return "door"; break; 
      case 5 : return "hive"; break; 
      case 6 : return "sticks"; break; 
      case 7 : return "heaven"; break; 
      case 8 : return "pate"; break; 
      case 9 : return "spine"; break; 
      case 10 : return "shin"; break; 

      default : return "VALUE NOT FOUND!!"; 
     } 
    } 
}