How to write a C Program to Count Vowels and Consonants in a String with an example?.
C Program to Count Vowels and Consonants in a String Example 1
This program allows the user to enter a string (or character array). Next, it will count how many numbers of Vowel and Consonant present in the user-specified string using If Else Statement.
/* C Program to Count Vowels and Consonants in a String */
#include <stdio.h>
int main()
{
char str[100];
int i, vowels, consonants;
i = vowels = consonants = 0;
printf("\n Please Enter any String : ");
gets(str);
while (str[i] != '\0')
{
if(str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u' ||
str[i] == 'A' || str[i] == 'E' || str[i] == 'I' || str[i] == 'O' || str[i] == 'U')
{
vowels++;
}
else
consonants++;
i++;
}
printf("\n Number of Vowels in this String = %d", vowels);
printf("\n Number of Consonants in this String = %d", consonants);
return 0;
}OUTPUT
ANALYSIS
First, we used While Loop to iterate every character in a String.
while (str[i] != '\0')
{Next, we used the If Else Statement to check whether each character is equal to a, e, i, o, u, A, E, I, I, O. And if it is TRUE, it is a Vowel otherwise, it’s a Consonant.
TIP: You can convert all the characters to the same case using strlwr or strupr, and then check for lowercase or uppercase elements in the If condition
if(str[i] == 'a' || str[i]== 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u' ||
str[i] == 'A' || str[i] == 'E' || str[i] == 'I' || str[i] == 'O' || str[i] == 'U') {
vowels++;
}str[] = Hello World
i = vowels = consonants = 0
While Loop First Iteration: while (str[ i ] != 0)
The condition is True because str[0] = H.
Within the While Loop, we used If Else Statement to check whether it is an Vowel or Consonant.
if(str[i] == ‘a’ || str[i]== ‘e’ || str[i] == ‘i’ || str[i] == ‘o’ || str[i] == ‘u’ || str[i] == ‘A’ || str[i] == ‘E’ || str[i] == ‘I’ || str[i] == ‘O’ || str[i] == ‘U’)
=> if(H == ‘a’ || H== ‘e’ || H== ‘i’ || H== ‘o’ || H== ‘u’ || H == ‘A’ || H == ‘E’ || H == ‘I’ || H == ‘O’ || H == ‘U’)
The above condition is false. So, it will execute the statement inside the Else block.
consonants++. It means consonants = 1
Second Iteration: while (str[ 1 ] != 0)
The condition is True because str[1] = e.
if(e == ‘a’ || e == ‘e’ || e == ‘i’ || e == ‘o’ || e == ‘u’ || e == ‘A’ || e == ‘E’ || e == ‘I’ || e == ‘O’ || e == ‘U’)
The above condition is True. So, it will execute the statement inside this block.
vowels++. It means vowels = 1
Do the same for remaining iterations
C Program to Count Vowels and Consonants in a String Example 2
This program allows the user to enter any string value and count the number of Vowel or Consonant in this string using ASCII Values.
/* C Program to count vowels and consonants in a String */
#include <stdio.h>
int main()
{
char str[100];
int i, vowels, consonants;
vowels = consonants = 0;
printf("\n Please Enter any String : ");
gets(str);
for(i = 0; str[i] != '\0'; i++)
{
if(str[i] == 97 || str[i] == 101 || str[i] == 105 || str[i] == 111 || str[i] == 117 ||
str[i] == 65 || str[i] == 69 || str[i] == 73 || str[i] == 79 || str[i] == 85)
{
vowels++;
}
else
consonants++;
}
printf("\n Number of Vowels in this String = %d", vowels);
printf("\n Number of Consonants in this String = %d", consonants);
return 0;
}OUTPUT
C Program to Count Vowels and Consonants in a String Example 3
This program is the same as the first example, but this time we used the Functions concept to separate the logic.
/* C Program to count vowels and consonants in a String */
#include <stdio.h>
int check_vowel(char);
int main()
{
char str[100];
int i, vowels, consonants;
vowels = consonants = 0;
printf("\n Please Enter any String : ");
gets(str);
for(i = 0; str[i] != '\0'; i++)
{
if(check_vowel(str[i]))
{
vowels++;
}
else
consonants++;
}
printf("\n Number of Vowels in this String = %d", vowels);
printf("\n Number of Consonants in this String = %d", consonants);
return 0;
}
int check_vowel(char c)
{
if (c >= 'A' && c <= 'Z')
c = c + 32;
if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
return 1;
return 0;
}OUTPUT
C Program to Count Vowels and Consonants in a String Example 4
In this program, we are using the Switch case (alternative to Else If) to count vowels and consonants
/* C Program to count vowels and consonants in a String */
#include <stdio.h>
int main()
{
char str[100];
int i, vowels, consonants;
vowels = consonants = 0;
printf("\n Please Enter any String : ");
gets(str);
for(i = 0; str[i] != '\0'; i++)
{
if( (str[i] >= 'a' && str[i] <= 'z') || (str[i] >= 'A' && str[i] <= 'Z') )
{
switch(str[i])
{
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
case 'A':
case 'E':
case 'I':
case 'O':
case 'U':
vowels++;
break;
default:
consonants++;
}
}
}
printf("\n Number of Vowels in this String = %d", vowels);
printf("\n Number of Consonants in this String = %d", consonants);
return 0;
}OUTPUT
C Program to Count Vowels and Consonants in a String Example 5
This program to count vowels and consonants is the same as the first example, but this time we are using the concept of the pointers.
/* C Program to count vowels and consonants in a String */
#include <stdio.h>
int main()
{
char str[100];
char *s = str;
int vowels, consonants;
vowels = consonants = 0;
printf("\n Please Enter any String : ");
gets(str);
while (*s)
{
if(*s == 'a' || *s == 'e' || *s == 'i' || *s == 'o' || *s == 'u' ||
*s == 'A' || *s == 'E' || *s == 'I' || *s == 'O' || *s == 'U')
{
vowels++;
}
else
consonants++;
s++;
}
printf("\n Number of Vowels in this String = %d", vowels);
printf("\n Number of Consonants in this String = %d", consonants);
return 0;
}OUTPUT
