Thread: Program to count how many of each letter has been typed

im trying to create a simple program since im just starting out ,that prints out a histogram telling how many of each letter has been typed in here's my program so far:

Code:

#include <stdio.h>
#define AMOUNT 47
int main()
{
    int c,b,i;
    int biggestnum = 0;
    int letters;
    int ndigit[AMOUNT];
    int numbers[AMOUNT];
    for( i = 0; i <= 47; i++)
        numbers[i] = 0;
    for( i = 0; i <= 23; i++)
    {
        for( letters = 'A'; letters <= 'Z' ; letters ++ )
        ndigit[i] = letters;
    }
    for ( i = 24; i <= 47; i++)
    {
        for( letters = 'a'; letters < 'z'; letters ++ )
        ndigit[i] = letters;
    }
    while((c = getchar()) != EOF )
    {
        for ( i = 0; i <= 47; i++)
        {
            if ( c == ndigit[i])
             ++numbers[i];
            if ( numbers[i] > biggestnum )
            biggestnum = numbers[i];
        }
    }
    for ( i = biggestnum; i > 0; --i )
    {
            printf("%4d   [" , i);
        for ( b = 0; b <= AMOUNT; b++ )
        {
            if (numbers[b] >= i)
            printf("x  ");
            else
            printf("   ");
        }
        printf("\n");
    }
}

ive been having some problems with it and wondered if anyone could help so far i havent included to the bottom part of the histogram as i cant get this section to work

ow thanks my maths has gone out the window -_-

how would you impliment using one array, that part confuses me

well first you'd dfine an array of 26 ie int array[26]; ,then im not to sure about the second bit

Do you understand the meaning of map in this context? To map a value to a location, you need to be able to choose other key values that will give up the data being stored.

yeah i think so thanks

Code:

#define AMOUNT 47
...
int numbers[AMOUNT];
...
for( i = 0; i <= 47; i++)
    numbers[i] = 0;

1. You're declaring an array with a named constant ("AMOUNT"). I'm guessing you're using the "for" loop to initialize all values of the array to zero. If this is the case, I would suggest using "AMOUNT" instead of a number constant in the comparison expression in your "for" loop. Therefore, if you make a change to the length of this array (such as laserlight recommends in post #2), you only have to change one line of code.

2. Arrays in C - You're declaring the "numbers" array to have 47 elements. The first index number starts at zero ("numbers[0]"), so the last index number would be "n-1" ("numbers[46]" in this case). Your "for" loop exceeds the bounds of this array by one (check your relational operator in the comparison expression).

ive tried all this but when i compile and type in a letter ,the screen just stays blank

Show your current code.

Cos I'm away from my computer atm I can't but basically I still have two arrays as when I attempted it one array confused me and as I'm still a beginner I decided to stick with what I know ,the only thing Ichanged was the array size to 52 And all the for loops that relate to the array size so that parts correct

here's my current code i changed it to one array and shortened it to only pick lower case characters im still having problems

Code:

#include <stdio.h>
#define AMOUNT 26
int main()
{
    int i,b,c;
    int currentnum,biggestnum;
    int numbers[AMOUNT];
    currentnum = biggestnum = 0;
    for ( i = 0; i < AMOUNT; ++i)
        numbers[i] = 0;
    while((c = getchar()) != EOF)
    {
            if( (c - 'a') < AMOUNT && (c - 'a') >= 0 )
            {
                ++numbers[c - 'a'];
                currentnum = numbers[c - 'a'];
                if (currentnum > biggestnum)
                biggestnum = currentnum;
            }
    }
    for ( i = biggestnum; i > 0; i--)
    {
        printf("%4d   ]",i);
        for ( b = 0; b > AMOUNT; b++)
            {
                if ( numbers[b] >= i )
                printf("x  ");
                else
                printf("   ");
            }
        printf("\n");
    }
}