C program for 5 function of vowels, cnt_words, reverse , C/C++ Programming

C Program for 5 FUNCTION OF VOWELS, CNT_WORDS, REVERSE

void input(char a[]);

void output(char a[]);

void reverse(char a[], char b[]);

char poli(char a[], char b[]);

int count(char a[], int l[]);

void vowels(char a[]);

void main()

{

          char a[30],b[30],j;

          int i,k=0,l[30];

          clrscr();

 

          for(i=0;i<30;i++)

          {

                   a[i]='0';

                   b[i]='0';

                   l[i]=0;

          }

          input(a);

          output(a);

          reverse(a,b);

          j=poli(a,b);

          k=count(a,l);

          vowels(a);

          printf("\n\nTHE NO OF WORDS IN STRING ARE %d\n\n",k);

          if(j=='e')

          {

                   printf("\n\nIT IS PALINDROME");

          }

          else

          {

                   printf("\n\nIT IS NOT PALINDROME");

          }

          getch();

}

 

 

void input(char a[])

{

          int i=0;

          printf("ENTER THE STRING= ");

          do

          {

                   a[i++]=getchar();

 

          }while(a[i-1]!='\n');

          a[i-1]='\0';

}

void output(char a[])

{

          int i=0;

          for(i=0;a[i]!='\0';i++)

          {

                   printf("%c",a[i]);

          }

          printf("\n\n");

}

void reverse(char a[],char b[])

 

{

          int i=0,j=0;

          i=strlen(a);

          i=i-1;

          for(;i!=-1;i--)

          {

                   b[i]=a[j++];

                   printf("%c",a[i]);

          }

}

char poli(char a[], char b[])

{

          int i=0;

          char j='e';

 

          for(i=0;a[i]!='\0';i++)

          {

                   if(a[i]!=b[i])

                   {

                             j='n';

                             break;

                   }

          }

          return(j);

}

int count(char a[], int l[])

{

          int i=0,k=0,c=0;

          for(i=0;a[i]!='\0';i++)

          {

                   if(a[i]==' ')

                   {

                             k++;  c++;

                   }

 

else

                             l[c]=l[c]+1;

          }

          c=1;

          printf("\n\n");

          for(i=0;l[i]!=0;i++)

          {

printf("THE NO OF CHARACTERS IN %d WORD ARE %d\n",c,l[i]);

                   c++;

          }

          k++;

          return(k);

}

void vowels(char a[])

{

          int j=0,a1=0,e=0,i=0,o=0,u=0,l[30],c=0;

          for(j=0;j<30;j++)

                   l[j]=0;

          for(j=0;a[j]!='\0';j++)

          {

                   if(a[j]=='a' || a[j]=='A')

                   {

                             a1++;

                             l[c++]=j+1;

                   }

                   if(a[j]=='e' || a[j]=='E')

                   {

                             e++;

                             l[c++]=j+1;

                   }

                   if(a[j]=='i' || a[j]=='I')

                   {

                             i++;

                             l[c++]=j+1;

                   }

                   if(a[j]=='o' || a[j]=='O')

                   {

                             o++;

                             l[c++]=j+1;

                   }

                   if(a[j]=='u' || a[j]=='U')

                   {

                             u++;

                             l[c++]=j+1;

                   }

          }

          printf("THE NO OF VOWELS \"a\" AND \"A\" ARE %d\n",a1);

          printf("THE NO OF VOWELS \"e\" AND \"E\" ARE %d\n",e);

          printf("THE NO OF VOWELS \"i\" AND \"I\" ARE %d\n",i);

          printf("THE NO OF VOWELS \"o\" AND \"O\" ARE %d\n",o);

          printf("THE NO OF VOWELS \"u\" AND \"U\" ARE %d\n",u);

}

 

OUTPUT :

ENTER THE STRING:

SHIVANI PRERNA SHRIDEVI VIRAL KAMLESH  

HSELMAK LARIV IVEDIRHS ANRERP INAVIHS

THE NO OF CHAR IN 1 WORD ARE 7

THE NO OF CHAR IN 1 WORD ARE 6

THE NO OF CHAR IN 1 WORD ARE 8

THE NO OF CHAR IN 1 WORD ARE 5

THE NO OF CHAR IN 1 WORD ARE 7

 

 

OUTPUT :

THE NO OF VOWELS A  2

THE NO OF VOWELS E  3

THE NO OF VOWELS  I  5

THE NO OF VOWELS O  0

THE NO OF VOWELS U  0

THE NO OF WORDS IN THE STRING 5

IT IS NOT PALINDROM

 

Posted Date: 9/10/2012 7:59:10 AM | Location : United States







Related Discussions:- C program for 5 function of vowels, cnt_words, reverse , Assignment Help, Ask Question on C program for 5 function of vowels, cnt_words, reverse , Get Answer, Expert's Help, C program for 5 function of vowels, cnt_words, reverse Discussions

Write discussion on C program for 5 function of vowels, cnt_words, reverse
Your posts are moderated
Related Questions
1. When developing this project in a Win32 Console Applications that includes the precompiled headers, enter in the Name: box, PRJ2[Your Full Last Name][Your First Initial] with no


This assignment is to be undertaken individually - no group work is permitted. Background information This assignment is an exercise in simple object-oriented programming and, acco

Explain each of the algorithms in a way that would be understandable to an intelligent person who is not familiar with programming. You should not use any code (or even pseudo code

We can combine more than one variable on the same line i.e.   float number1,number2,number3; etc Sometimes we want to mix the variable types used on the same line, this could

3. Write a program to encrypt and decrypt strings of characters using the following ciphers: a) Caesar cipher b) Vigenere cipher c) Matrix transposition cipher Your program shoul

Writing and compiling a program from a given Use Case definition. Follow the Average Temperature in Paradise

Byteland county is very famous for luminous jewels. Luminous jewels are used in making beautiful necklaces. A necklace consists of various luminous jewels of particular colour. Nec

One person who is specialist at programming and solving problems with a computer Project Description: Potential computer, hardware, programming and software genius, I look

Explain Object-oriented programming Object-oriented programming (OOP) attempts to meet these requirements, providing methods for managing enormous complexity, achieving reuse o