C Exercises: Count the number of vowels and consonants
C Pointer : Exercise-13 with Solution
Write a program in C to count the number of vowels and consonants in a string using a pointer.
Pictorial Presentation:
Sample Solution:
C Code:
#include <stdio.h>
int main()
{
char str1[50];
char *pt;
int ctrV,ctrC;
printf("\n\n Pointer : Count the number of vowels and consonants :\n");
printf("----------------------------------------------------------\n");
printf(" Input a string: ");
fgets(str1, sizeof str1, stdin);
//assign address of str1 to pt
pt=str1;
ctrV=ctrC=0;
while(*pt!='\0')
{
if(*pt=='A' ||*pt=='E' ||*pt=='I' ||*pt=='O' ||*pt=='U' ||*pt=='a' ||*pt=='e' ||*pt=='i' ||*pt=='o' ||*pt=='u')
ctrV++;
else
ctrC++;
pt++; //pointer is increasing for searching the next character
}
printf(" Number of vowels : %d\n Number of consonants : %d\n",ctrV,ctrC-1);
return 0;
}
Sample Output:
Pointer : Count the number of vowels and consonants : ---------------------------------------------------------- Input a string: string Number of vowels : 1 Number of consonants : 5
Flowchart:
C Programming Code Editor:
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C Programming: Tips of the Day
Static variable inside of a function in C
The scope of variable is where the variable name can be seen. Here, x is visible only inside function foo().
The lifetime of a variable is the period over which it exists. If x were defined without the keyword static, the lifetime would be from the entry into foo() to the return from foo(); so it would be re-initialized to 5 on every call.
The keyword static acts to extend the lifetime of a variable to the lifetime of the programme; e.g. initialization occurs once and once only and then the variable retains its value - whatever it has come to be - over all future calls to foo().
Ref : https://bit.ly/3fOq7XP
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