猪拉丁语翻译（句子）

Question

我正在制作一个猪拉丁语翻译，我真的不知道从哪里去。我有基本的代码，但我需要修改它以便翻译整个句子。如果有人能告诉我如何使用我的字符串[]字我真的很感激它。非常感谢！

import java.io.*; 
import java.util.*; 
import java.util.Arrays; 

public class Main 
{ 
    public static void main (String[] args) 
    { 
    System.out.print("Please enter a phrase to translate: "); 
    Scanner scan = new Scanner(System.in); 
    String str = scan.nextLine(); 
    String[] words = str.split("\\s+"); 
    String answer = ""; 
    if (str.startsWith("a") || str.startsWith("e") || str.startsWith("i") || str.startsWith("o") || str.startsWith("u")) 
    { 
     System.out.print(str + "way"); 
     } 
    else 
     { 
     answer = str.substring(2,str.length()); 
     String answer2 = str.substring(1,str.length()); 
     String answer3 = str.substring(3,str.length()); 
     String answer4 = str.substring(4,str.length()); 
     String d = str.substring(0,4); 
     if (!(d.contains("a") || d.contains("e") || d.contains("i") || d.contains("o") || d.contains("u"))) 
      { 
      System.out.print(answer4 + d + "ay"); 
      } 
     else 
     { 
      String c = str.substring(0,3); 
      if (!(c.contains("a") || c.contains("e") || c.contains("i") || c.contains("o") || c.contains("u"))) 
      { 
      System.out.print(answer3 + c + "ay"); 
      } 
      else 
      { 
       String b = str.substring(0,2); 
       if (!(b.contains("a") || b.contains("e") || b.contains("i") || b.contains("o") || b.contains("u"))) 
       { 
        System.out.print(answer + b + "ay"); 
        } 
       else 
       { 
        String a = str.substring(0,1); 
        if (!(a.contains("a") || a.contains("e") || a.contains("i") || a.contains("o") || a.contains("u"))) 
        { 
         System.out.print(answer2 + a + "ay"); 
         } 
       } 
      } 
     } 
    } 
} 
} 

Answer 1

这应该是阵列中的每个单词翻译，只需使用一个for循环你的整个周围，如果块

for(String word: words) { 
    //put your whole if block here 
    //word is the variable to translate (i.e. word.startsWith("a")) 
} 

或者初学者：

for(int i = 0; i < words.length(); i++) { 
    //put if statements here 
    //words[i] is the variable to translate (i.e. words[i].startsWith("a")) 
} 

你也需要改变所有str循环内的引用为word[i]，因为那是您想要使用的变量。

Answer 2

你可以有一个for循环，通过每个字去，和你的逻辑应用到每个字：

for(String str : words) { 

    if (str.startsWith("a") || str.startsWith("e") || str.startsWith("i") || str.startsWith("o") || str.startsWith("u")) 
    { 
     System.out.print(str + "way"); 
    } 
    else 
    { 
     answer = str.substring(2,str.length()); 
     String answer2 = str.substring(1,str.length()); 
     String answer3 = str.substring(3,str.length()); 
     String answer4 = str.substring(4,str.length()); 
     String d = str.substring(0,4); 
     if (!(d.contains("a") || d.contains("e") || d.contains("i") || d.contains("o") || d.contains("u"))) 
     { 
      System.out.print(answer4 + d + "ay"); 
     } 
     else 
     { 
      String c = str.substring(0,3); 
      if (!(c.contains("a") || c.contains("e") || c.contains("i") || c.contains("o") || c.contains("u"))) 
      { 
       System.out.print(answer3 + c + "ay"); 
      } 
      else 
      { 
       String b = str.substring(0,2); 
       if (!(b.contains("a") || b.contains("e") || b.contains("i") || b.contains("o") || b.contains("u"))) 
       { 
        System.out.print(answer + b + "ay"); 
       } 
       else 
       { 
        String a = str.substring(0,1); 
        if (!(a.contains("a") || a.contains("e") || a.contains("i") || a.contains("o") || a.contains("u"))) 
        { 
         System.out.print(answer2 + a + "ay"); 
        } 
       } 
      } 
     } 
    } 
} 

Answer 3

您可以使用循环增强在你如果块整个数组翻译：

for(String word: words) { 
    /* 
    *if block goes here 
    *... 
    */ 
} 

虽然创建这样的翻译是相当不安全的（例如代码注入），除非输入是从可信源放英寸

如果您熟悉词法分析器/解析器，则应使用生成器（如ANTLR）来创建更安全（更复杂更复杂）的语言翻译器。

Answer 4

#include stdio.h> 
#include conio.h> 
#include stdlib.h> 
#include string.h> 

void main() 
{ 
     char * sUserInput ; 
     char * check  ; 
     char * firstletter; 

     char ch = NULL; 
     int  i   = 0 ; 
     int  j   = 0 ; 
     int  k   = 0 ; 
     int  spacecount = 0 ; 
     int  count  = 0 ; 
     int  count1  = 0 ; 
     int  var  = 0 ; 
     int  gold  = 1 ; 

     sUserInput = (char *) malloc(sizeof(char)); 
     check  = (char *) malloc(sizeof(char)); 
     firstletter = (char *) malloc(4*sizeof(char)); 

     printf("Enter the sentence \n"); 

     while(ch != '\n') 
     { 
      scanf("%c",&ch); 
      sUserInput[i]=ch; 
      sUserInput = (char *) realloc(sUserInput,((i+2))*sizeof(char)); 
      i++; 
      if(ch == ' ') 
      { 
       spacecount++; 
      } 
     } 

     while(sUserInput[j] != '\n') 
     { 
      count1++; 
      if(gold==1) 
      { 
        firstletter[var]=sUserInput[j]; 
        firstletter[var+1]='a'; 
        firstletter[var+2]='y'; 
        gold=0; 
        j++; 
      } 
      if(sUserInput[j] !=' ') 
      { 
        check[k] = sUserInput[j]; 
        check = (char *) realloc(check,((k+2))*sizeof(char)); 
        printf("%c",check[k]); 
        k++; 
      } 
      if(sUserInput[j] ==' ' || (count1 == (i-(spacecount+2)))) 
      { 
        for(count=0;count<3;count++) 
        { 
          printf("%c",firstletter[count]); 
        } 
        var=0; 
        gold=1; 
        printf(" "); 
      } 
      j++; 
     } 

     free(sUserInput); 
     free(check); 
     free(firstletter); 
     getch(); 
}