Using pointers, write a program in ‘C’ to count the occurrence of each character in a given string. 10m Dec2006
#include <stdio.h>
#include <string.h>
void main()
{
char string[100],*ptr;
int c=0,count[26]={0};
clrscr();
printf(“Enter a string\n”);
gets(string);
ptr=string[0];
while (*ptr)
{
/* Consider characters from ‘a’ to ‘z’ only */
if ( ptr >= ‘a’ && ptr <= ‘z’ )
count[string[c]-‘a’]++;
c++;
ptr++;
}
for(c=0;c<26;c++)
{
if(count[c]!=0)
printf(“%c occurs %d times in string.\n”,c+’a’,count[c]);
}
getch();
}
[codesyntax lang=”c”]
#include <stdio.h> #include <string.h> void main() { char string[100],*ptr; int c=0,count[26]={0}; clrscr(); printf("Enter a string\n"); gets(string); ptr=string[0]; while (*ptr) { /* Consider characters from 'a' to 'z' only */ if ( ptr >= 'a' && ptr <= 'z' ) count[string[c]-'a']++; c++; ptr++; } for(c=0;c<26;c++) { if(count[c]!=0) printf("%c occurs %d times in string.\n",c+'a',count[c]); } getch(); }
[/codesyntax]
Screen Shots: