Find character frequency in a string - c

Question

Im trying to find character frequency in a string,i wrote the following code,but it does'nt show any output.All im trying is to fill the character array with respective counts.
When i tried to debug,it some how gives output,but prints some garbage value.
#include<stdio.h>
/* Program for printing character frequency in string */
charcount(char *,int *);
int main()
{
int n,i=0;
printf("Enter n :");
scanf("%d",&n);
char var[n];
int count[100]; // array for storing character frequency
printf("Enter string :");
fflush(stdin);
scanf("%s",var);
charcount(var,count); // calling frequeny function
for(i=0;i<strlen(count);i++)
{
printf("%d\n",count[i]);
}
getch();
return 0;
}
charcount(char *p,int *q)
{
for(;*p;p++)
{
q[*p]++;
}
}

Answer 1

You have few problems in your code:
count array is not initialized.
You are applying strlen() on an integer array.
count array should be 256 (count[256]) to cover all possible ascii chars. For example, if your input is abcd you'll go out of bound of array as d is 100.
You are printing the wrong count:
printf("%d\n",count[i]); should be printf("%d\n",count[var[i]]);
Declare proper prototype for charcount().
After fixing these:
#include<stdio.h>
/* Program for printing character frequency in string */
void charcount(char *,int *);
int main()
{
int n,i=0;
printf("Enter n :");
scanf("%d",&n);
char var[n];
int count[256]={0}; // array for storing character frequency
printf("Enter string :");
scanf("%s",var);
charcount(var,count); // calling frequeny function
for(i=0;i<strlen(var);i++)
{
printf("%d\n",count[var[i]]);
}
return 0;
}
void charcount(char *p,int *q)
{
for(;*p;p++)
{
q[*p]++;
}
}
Make sure to compile in C99 or C11 (e.g. gcc -std=c99 file.c) mode as VLAs are not supported in earlier standards of C.

Answer 2

You need to initialize your count array. Otherwise it will have garbage values in it by default. You can initialize the whole array to 0 like so:
int count[100] = {0};

Answer 3

Your count array may not be large enough to hold all printable values (even assuming ASCII), and it should be 0 initialized. Your for loop should be checking against the length of var, not count, since you cannot sensibly treat the count integer array as a string.
int count[1<<CHAR_BIT] = {};
/*...*/
for(i=0;i<strlen(var);i++)
{
printf("%d\n",count[var[i]]);
}
Well, it really depends on what you want to output. If you intend to output all of count, then:
for(i=0;i<sizeof(count)/sizeof(count[0]);i++)
{
printf("%d\n",count[i]);
}

Answer 4

int count is nothing but a hashmap
Your code will not work for this string "abcd"
count['a'] = val // Works fine ASCII value of a is 97
count['b'] = val // Works fine ASCII value of a is 98
count['c'] = val // Works fine ASCII value of a is 99
count['d'] = val ; // Undefined Behaviour ASCII value of d is 100
The size should be equal to ASCII set length
int count[128] = {};

Answer 5

JAVA program to print "*" as much times as the occurrence of a character in String.
Or char_in_String : frequency of char_in_String
Instead of * you can print frequency count
public class CharFreq
{
public static void main(String[] args)
{
String s = "be yourself "; System.out.println(s);
int r=0;
char[] str = s.toCharArray();
for (int i = 0; i < str.length; i++)
{
int cnt = 0;
if (str[i] != ' ')
{
for (int j = 0; j < str.length; j++)
{
if (str[i] == str[j])
{
cnt++; r=j;
}
}
if(i==r)
{
System.out.print(str[i] + ":");
for (int k = 1; k <=cnt; k++)
System.out.print("*");
System.out.println();
}
}
}
}
}
Output:
be yourself
b:*
y:*
o:*
u:*
r:*
s:*
e:**
l:*
f:*

Answer 6

Your count[100] is not large enough. Assume you only enter "A - Z" or "a - z" it's still not large enough because 'z' is 122, then your charcount will increase count[122].
You should consider change int count[100] to int count[128] = { 0 }