array either doesnt print out a count or it prints out multiple times

Question

this small program basically should get user input, read if it has any a's or A's and store those counts into Array[]. then print out that count of Array[].

#include<stdio.h>

main()
{
    int Array[100] = {0};

    int i;

    char input[100];

    printf("Enter a message:...");
    gets(input);

    for(i=0; input[i]!='\0'; i++)
    {
        if (input[i]=='a' || input[i]=='A') {Array[i]++;}   
    }

    printf("Total A's:%d\n",Array[i]);
}

example input/output from console:

 Enter a message:...harry 

 Total A's:0

clearly there is an 'a' in the input, but its not being considered. now i tried putting the printf in the for loop, but its would output this:

Enter a message:...harry 

 Total A's:0 

 Total A's:1 

 Total A's:0 

 Total A's:0 

 Total A's:0

what's happening is its apparently reading how many chars is the message, and checking for each index if it has a, which is supposed to do, but it's also printing out at each index this output.

Its supposed to only printf ONCE.so putting printf in loop does work, but generates this "bug" with it.

Answer 1

The problem is

 printf("Total A's:%d\n",Array[i]);

does not do what you think it does. The value you're trying to print is useless, from the point of logic.

Instead of incrementing the count for each occurrence of 'A' or 'a', you're just incrementing the initial value of the particular index ( of the Array ) for which you've found a match in the input. After that, you're printing the value of the variable of the index, for which the value has not been altered, in any way (hint: check the value of i after the loop).

Apparently, you don't need an array to hold the count of a single variable. Just use a normal variable, initialize it to 0 and keep on incrementing it.

Answer 2

At the last iteration i equals 99. Then i is incremented to 100 so the loop ends and then you want to print Array[100]. That index is out of bounds so it is undefined behavior.

As you say in the comment, you want to store the count of all characters in the array. Then calculate the letter to an index in the array based on its position in the alphabet.

#include <ctype.h>

for(i=0; input[i]!='\0'; i++)
{
  char c = tolower(input[i]);
  if (isalpha(c))
   Array[c - 'a']++;
}

This way 'a' will be index 0, 'b' 1 etc

'a' is an integer value (97 in the ascii table). 'b' is 98, c = '99' etc. So when you have 'a' in a variable and you subtract a literal 'a' the result = 0. 'b' - 'a' = 1 'c' - 'a' = 2 etc

Check here for ascii table values:

http://www.asciitable.com/

Answer 3

apparently, i just needed to use another variable j and change index in printf to 0.

#include<stdio.h>

main()
{
    int Array[100] = {0};

    int i;

    char input[100];

    printf("Enter a message:...");
    gets(input);

    for(i=0; input[i]!='\0'; i++)
    {   int j = 0;
        if (input[i]=='a' || input[i]=='A') {Array[j]++;}   
    }

    printf("Total A's:%d\n",Array[0]);
}

SOLVED :)

Answer 4

Your problem is a classic frequency problem. Whenever you need to calculate the frequency of a set of values, you use an array capable of holding that number of values. In your case, you need to know the frequency of the occurrence of 'a's and 'A's for a total of 2 elements (initialized to zero).

As you execute your code and test for the occurrence of any one of the values, you simply increment the element that corresponds to that value, e.g.

a_array[0]++;   /* when an 'a' is encountered */
a_array[1]++;   /* when an 'A' is encountered */

After you scan the entire string, the number of 'a's are in a_array[0] and the number of 'A's in a_array[1].

Putting that together, along with replacing gets (which you NEVER, EVER USE AGAIN) with fgets, you can do something like:

#include <stdio.h>
#include <string.h>

#define MAXC 256

int main (void) {

    int a_array[2] = {0};   /* if you are counting 'a's and 'A's, you need 2 */
    char buf[MAXC] = "";

    while (fgets (buf, MAXC, stdin)) /* for each line of input */
    {
        char *p = buf;
        size_t len = strlen(p);     /* get the length */

        if (buf[len-1] == '\n')     /* remove the trailing '\n' */
            buf[--len] = 0;         /* by overwriting with '\0' */

        for (; *p; p++)             /* for each char in buf */
            if (*p == 'a')          /* is it an 'a'? */
                a_array[0]++;       /* if so increment the 'a' counter */
            else if (*p == 'A')     /* if it is an 'A' */
                a_array[1]++;       /* increment the 'A' counter */

        printf ("\nthere are '%d' a's and '%d' A's in\n'%s'\n",
                a_array[0], a_array[1], buf);

        a_array[0] = a_array[1] = 0;    /* zero for next line */
    }

    return 0;
}

Example Use/Output

$ printf "my dog has A lot of fleas\nA snake has none - Amen.\n" | \
./bin/num_aAs 

there are '2' a's and '1' A's in
'my dog has A lot of fleas'

there are '2' a's and '2' A's in
'A snake has none - Amen.'

Now typically, you will not hardcode a_array[0] and a_array[1], but will use a logical index to hold the values. For example, to calculate the frequency of each occurrence of each letter a-z (26 characters in all), you could use:

int c_array[26] = {0};

Then when testing for each character, you simply increment c_array[*p - 'a']++;, e.g.

while (fgets (buf, MAXC, stdin))     /* for each line of input */
{
    char *p = buf;
    /* trim '\n' */
    for (; *p; p++)                 /* for each char in buf */
        if ('a' <= *p && *p <= 'z')
            c_array[*p - 'a']++;
    ...

The resulting c_array will hold the number of 'a's in c_array[0] and each corresponding letter frequency in the subsequent element through c_array[25] which hold the frequency for all 'z's.

For example, if you wanted to experiment with a logical index scheme for your problem, you could use something on the order of:

    for (; *p; p++)                     /* for each char in buf */
        if (*p == 'a')                  /* is it an 'a'? */
            a_array[*p - 'a']++;        /* if so increment the 'a' counter */
        else if (*p == 'A')             /* if it is an 'A' */
            a_array[*p - 'A' + 1]++;    /* increment the 'A' counter */