查找字符串中的字符频率

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:

1. count array is not initialized.

2. You are applying strlen() on an integer array.

3. 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.

4. You are printing the wrong count:

   printf("%d\\n",count[i]); should be printf("%d\\n",count[var[i]]);

5. 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 (eg 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

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 4

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 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 }