Question

Using C++ Create a program that asks the user to input a string value and then outputs the string in the Pig Latin form.

- If the string begins with a vowel, add the string "-way" at the end of the string. For “eye”, it will be “eye-way”.

- If the string does not begin with a vowel, first add "-" at the end of the string. Then rotate the string one character at a time; that is, move the first character of the string to the end of the string until the first character of the string becomes a vowel. Then add the string "ay" at the end. For example, the Pig Latin form of the string "There" is "ere-Thay".

- Strings such as "by" contain no vowels. In cases like this, the letter y can be considered a vowel. So, for this program, the vowels are a, e, i, o, u, y, A, E, I, O, U, and Y. Therefore, the Pig Latin form of "by" is "y-bay".

- Strings such as "1234" contain no vowels. The pig Latin form of the string "1234" is "1234-way". That is, the pig Latin form of a string that has no vowels in it is the string followed by the string "-way". The program should store each character of a string into a linked list and use the rotate function, which removes the first item and puts it at the end of the linked list. The linked list should be implemented as a generic type.

Answer 1

• #include <stdio.h>
  #include <string.h>

  
  // this function returns 1 if first character of the string is vowel
  // and returns 0 if first character is not vowel
  int isFirstCharVowel(char *s){
     // Check if first character is vowel
     if (s[0]=='a' || s[0]=='e' || s[0]=='i' || s[0]=='o' || s[0]=='u' || s[0]=='y' || s[0]=='A' || s[0]=='E' || s[0]=='I' || s[0]=='O' || s[0]=='U' || s[0]=='Y')
     {
         return 1;
     }
     return 0;
  }

  // this function returns 1 if any character of the string is vowel
  // and returns 0 if none of the character is vowel
  int isVowelPresent(char *s){
     int length=strlen(s);
     // Traverse the string to check vowel at every position
     for (int i = 0; i < length; ++i)
     {
         // If vowel is present return true
         if (s[i]=='a' || s[i]=='e' || s[i]=='i' || s[i]=='o' || s[i]=='u' || s[i]=='y' || s[i]=='A' || s[i]=='E' || s[i]=='I' || s[i]=='O' || s[i]=='U' || s[i]=='Y')
         {
             return 1;
         }
     }
     // If we have reached this point means vowels is not present hence return 1
     return 0;
  }

  // This function rotates string by one character
  void rotate(char *s){
     int length=strlen(s);
     char first=s[0];
     for (int i = 0; i < length-1; ++i)
     {
         s[i]=s[i+1];
     }
     s[length-1]=first;
  }

  int main(int argc, char const *argv[])
  {
     char s[100];
     printf("Please enter the string.\n");
     gets(s);
     // Calculate lenth of the string using strlen() function
     int length=strlen(s);

     // Check if the first character is vowel
     if (isFirstCharVowel(s))
     {
         s[length]='-';
         s[length+1]='w';
         s[length+2]='a';
         s[length+3]='y';
         s[length+4]='\0';
     }
     // If first character is not vowel
     else{
         // Check if string contains vowel at any position
         if (isVowelPresent(s))
         {
             // add - at the end of string
             s[length]='-';
             s[length+1]='\0';
             while(!(s[0]=='a' || s[0]=='e' || s[0]=='i' || s[0]=='o' || s[0]=='u' || s[0]=='y' || s[0]=='A' || s[0]=='E' || s[0]=='I' || s[0]=='O' || s[0]=='U' || s[0]=='Y')){
                 rotate(s);
             }
             length=strlen(s);
             // Add ay at the end of string
             s[length]='a';
             s[length+1]='y';
             s[length+2]='\0';
         }
         // If none of the string's character is vowel
         else{
             // Add -way at the end of string
             s[length]='-';
             s[length+1]='w';
             s[length+2]='a';
             s[length+3]='y';
             s[length+4]='\0';
         }
     }

     printf("%s\n", s);
    
     return 0;
  }