Comparing ASCII value of input (C Program) - c

Question

For practice I am trying to create a program that counts the number of each digit, alphabetic character, and "whitespace" character in a line.
My program is printing fine however my counter is not correctly adding each character it reads. Below is the code for the program (I've only started learning about a week ago so please excuse any glaring issues).
I believe the my main issue rests in the if/else if statements where I compare the int c to a variety of ASCII values.
/* Print Count of Each Char, Spaces, and Digits */
int main(void) {
int c, i, CountWhitespace, count, Alpha;
int CountCharacter[26];
int CountDigits[10];
CountWhitespace = 0;
for (i = 0; i < 10; ++i) {
CountDigits[i] = 0; }
for (i = 0; i < 26; ++i) {
CountCharacter[i] = 0; }
while ((c = getchar()) != '\n') {
if (c >= '0' && c <= '9') {
++CountDigits[c]; }
else if (c == '\t' || c == ' ') {
++CountWhitespace; }
else if (c >= 'a' || c <= 'z') {
++CountCharacter[( c - 'a')]; }
else if (c >= 'A' && c <= 'Z') {
++CountCharacter[(c - 'A')]; }
}
printf("Whitespace Characters: %d\n", CountWhitespace);
for (i = 0; i < 10; i++) {
printf("%d appears %d times.\n", i, CountDigits[i]); }
for (Alpha = 'a', count = 0; count < 26; count++, Alpha++) {
printf("%c appears %d times.\n", Alpha, CountCharacter[count]); }
return EXIT_SUCCESS;
}
Below is an example of the output:
The green characters are the input, and as you may be able to tell, none of them are added to their corresponding variables.

Answer 1

You need to fix the digits counter:
if (c >= '0' && c <= '9') {
++CountDigits[c];
}
This will be increment CountDigits[48] (for '0'), etc. Not what you want. You've got it right for a..z; do the same here:
if (c >= '0' && c <= '9') {
++CountDigits[c - '0'];
}