1)Write a function CountNumberOfOccurrence which take a string with maximum size of 49
characters (char array size is 50) and returns the number of occurrence of English characters,
spaces, digits and other characters. The number of occurrence for English characters, digits, and
other characters using call by reference and the return value gives the number of occurrence for
spaces. In your main program, you ask the user to input a string, call CountNumberofOccurrence
function to get the corresponding number of occurrence of English characters, spaces, digits and
other characters and display the results. A sample input/output looks like following:
Input:
Enter a string: 8djfa -99jdfadf 882 adf *
Output:
Number of English characters = 13
Number of space = 3
Number of digits = 6
Number of other characters = 2

so, whats your question ?

@ poster this might help you.

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>

void Count_Number_of_occurrence(int *Engchar1,int *space1,int *digit1, int *other1);
void print_result(int *Engchar2,int *space2,int *digit2, int *other2);
int main()
{
int Engchar,space,digit,other;


Count_Number_of_occurrence(&Engchar,&space,&digit,&other);
print_result(&Engchar,&space,&digit,&other);


_getche();
return 0;
}
void Count_Number_of_occurrence(int *Engchar1,int *space1,int *digit1, int *other1)
{
int Eng,spac,dig,oth;
Eng=0;
spac=0;
dig=0;
oth=0;
char ch;
int loop=0;
printf("\n Please enter a statement \n";
while(((ch=(char) getche())!='\0')&&(loop<50))
{
if(ch==' ')
spac++;
else if(ch>='0'&&ch<='9')
dig++;
else if(((ch>='a')&&(ch<='z'))||((ch>='A')&&(ch<='Z')))
Eng++;
else
oth++;

loop++;
}
*Engchar1=Eng;
*space1=spac;
*digit1=dig;
*other1=oth;
}

void print_result(int *Engchar2,int *space2,int *digit2, int *other2)
{
printf("\n Number of English Characters = %d\n",*Engchar2);
printf("\n Number of Space = %d\n",*space2);
printf("\n Number of digits = %d\n",*digit2);
printf("\n Number of other characters = %d\n",*other2);
}