to counting the number of the occurrences in the input data and sort it in alphabet

[sodens48]

Hi, i need help on these question..i already search for the answer on the google and still not one of that help..i need to write a c program to count the occurrence of the input data and display it in alphabetical order.. for example i insert DSADSSAA and it need to show A=3,D=2,S=3..
so,can one of the expert help me..even the hint on how to count it also help me a lot..

[weaknessforcats]

Use an array for the 26 counts.

From your ASCII table you can see that an 'A' is 65.

So read 1 character of input.
Subtract 65
Now you have the element number in he array.
Increment the count in that position.

So 'A' is 65. You subtract 65. Your result is 0 so increment array[0].
So 'B' is 66. You subtract 65. Your result is 1 so increment array[1].

Off you go.

[sodens48]

hi weaknessforcats..
thx for for the help..right now i'm try to do your suggestion..and before that,i manage do the program,but it seem the program will be to long if i want to check and count the alphabet from a-z..the program is like below..

#include <stdio.h>
#define MAXNUM 15

int countchar (char []);
int main()

{
char message [MAXNUM];
int numchar;
int i, count = 0;
printf("User key in: ");
gets(message);
countchar (message);
getch();

} // END OF MAIN

//FUNCTION IMPLEMENTATION

int countchar (char list[])
{
int input,a=0,b=0,c=0,d=0,e=0,f=0, count = 0;
for (input = 0; list[input] != '\0'; input++)
{
if(list[input]=='a')
++a;
if(list[input]=='b')
++b;
if(list[input]=='c')
++c;
if(list[input]=='d')
++d;
if(list[input]=='e')
++e;
if(list[input]=='f')
}
if(a!=0)
printf("A = %d\n", a);
if(b!=0)
printf("B = %d\n", b);
if(c!=0)
printf("C = %d\n", c);
if(d!=0)
printf("D = %d\n", d);
if(e!=0)
printf("E = %d\n", e);
if(f!=0)
printf("F = %d\n", f);
return (count);
}

so,is there any way to make it more simple??

[weaknessforcats]

The idea is to calculate the array element to increment.

Just follow my steps:

Read one character
Subtract 65.
Increment the array element at that position.

If you are using capital and lower case letters then there are 52 counts:

int counts[52];
 

You need to set these counts to zero.

So if you read a char then you would:

++counts[char_I_read - 65];
 

and that's it. You do not need to test for the particuar character.

The array is a map. The first 26 elements are for A to Z and the last 26 are for a to z.

When you display the counts you can use this mapping to determine what letter to display. In this case ADD 65 to the array position:

So if you want to display counts[0] you can set a char to 0, then add 65 and wen you display the char you will see A.

Therefore in your display loop all you need is:

printf("%c %d\n", (char)(i+65+), counts[i]); 

[sodens48]

It seem i still cant do your method..i have been struggling for over a week and still cant do like you suggest..can you give me other example please..i'm really desperate right now..

[weaknessforcats]

OK. Here's the main():

int main()
{
char input[80];   //the data entered
int counts[26];   //[0] is for A. [1] is for B, etc
 
InitializeCounts(counts, 26);
 
cout << "Enter your letters\n" << endl;
cin >> input;
 
AnalyzeData(input, counts);
DisplayCounts(counts, 26);
}

This solution just has the cuonts array. There are three functions. One function initializes the counts array elemets to 0. The second takes the inout and processes it to make he counts and the third displays the counts.

Note that an A has a value of 65. Moving 65 to a char and displaying the char will produce an A on the screen. The display function uses this fact to generate the letter to display based on the counts element position.

Here are the three functions:

void AnalyzeData(char* data, int* counts)
{
      for(int i = 0; data[i] != '\0'; ++i)
      {
           ++counts[data[i] - 65];
      }
 
 
}
void DisplayCounts(int* counts, int numelements)
{
    char temp;
          for (int i = 0; i<numelements; ++i)
          {
                    temp = i +65;   //create letter to display
                    cout  << temp << "  " << counts[i] << endl;
          }
}
void InitializeCounts(int* counts, int numelements)
{
 
          for (int i = 0; i<numelements; ++i)
          {
                    counts[i] = 0;
          }
}

[donbock]

Another approach is to first sort the array; and then scan it from front to back, counting the length of each sequence of identical values. With this approach you don't need to make any assumptions about the nature of your input.

However, sometimes you want to make assumptions. For example, do you intend to treat 'A' and 'a' as different tokens? If not, then something more is needed. It might be as simple as using toupper() to convert all characters to uppercase in the sorting step.