int main()
{
char a[100],b;
int i = 0,n = 0,len = 0;
printf("Enter The String Of Characters : ");
scanf("%s",a);
printf("\n\nEnter The Character You want to Count : ");
scanf("%c",&b);
len = strlen(a);
fflush(stdin);
for(i = 0;i <= len;i++)
{
if(b == a[i])
{
n++;
}
}
printf("\n\nNumber Of Occurences of %c in %s is : %d",b,a,n);
getch();
return 0;
}
int main()
{
char string[80];
char ch;
int i, count;
cout << "Enter a character: ";
cin >> ch;
cout << "Enter a string: ";
cin >> *string;
count(string, ch);
return 0;
}
int count(const char * const str, char a)
{
int counter = 0;
for (int i = 0; str[i] != ‘\0’; i++)
if (str[i] == a)
*counter++;
cout << "Number of occurrences of " << a << " in " << str << " = " << count << endl;
return counter;
}
@kpr07
You can use Hash array for this problem because string may have only 255 different type of character
#include<stdio.h>
#include<string.h>
int main()
{
char st[100],c;
int Hash[256] = {0};
scanf("%s",st);
fflush();
scanf("%c",c);
int l = strlen(st);
for(i=0;i<l;i++)
Hash[st[i]]++;
printf("%d\n",Hash[c]); // this will give you the occurrence of character in a string.
return 0;
}