Using getchar() to store a value to compare with an integer

Question

i'm trying to count how many times a user inputs a certain digit and assign total number of instances to a location in a 10 row array (0-9). For instance if the user inputs 888, it will assign 3 to location arr1[8].

#include <stdio.h>
#include <stdlib.h>

int main(void){
    int arr1[10] = {0};
    int c;

    while ((c = getchar()) != '\n'){
        for (int i = 0; i <= 9; i++){
            if (c == i) // This isn't doing what I want it to do
                arr1[i] += 1;
        }
    }
    for (int i = 0; i <= 9; i++)
    printf ("%c ", arr1[i]);
}

The trouble seems to be the line that i've added to comment above. The line if (c == i) is intended to compare a user inputed digit (as it's entered by the user, not the ASCII value) and compare it with i.

So how can I compare c and i as the same type? I've tried changing the type of getchar() to char, signed int, unsigned int but it doesn't seem to make any difference.

have you looked at standard runtime function `isdigit()`? ctype.h — AndersK, May 13 '13 at 12:21
possible duplicate of [How to convert char to integer in C?](http://stackoverflow.com/questions/868496/how-to-convert-char-to-integer-in-c) — Shafik Yaghmour, May 13 '13 at 12:26

score 2 · Accepted Answer · answered May 13 '13 at 12:25

You have to substract '0'.
You are printing char using %c for count.
You are incrementing by the count by i in the loop rather you wanted to increment by 1 for each character.

Corrected code:

    int main(void){
        int arr1[10] = {0};
        int c;
        while ((c = getchar()) != '\n'){
            for (int i = 0; i <= 9; i++){
                if (c - '0'  == i) 
                    arr1[i] += 1;
            }
        }
        for (int i = 0; i <= 9; i++)
        printf ("%d ", arr1[i]);
    }

yeah sorry, the `=+ i` was just a typo when I rewrote a simpler version for this question. — Martin, May 13 '13 at 12:34

score 1 · Answer 2 · answered May 13 '13 at 12:20

1

You have to perform the following operation to convert the ASCII value to corresponding integer,

c=c-48;

Inside your for loop,

arr1[i] += i;

should be,

arr1[i] += 1;

answered May 13 '13 at 12:20

Deepu

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Stefano Sanfilippo · Answer 3 · 2013-05-13T12:29:18.467

getchar will return ASCII code for c, not the numerical value. Simplest solutions exploits the fact that numbers 0-9 are sequential in the ASCII table, just subtract the offset:

int number = getchar() - '0';
if (number == i) {
    //....
}

This is just an example, always check the return value of getchar for error codes.

Finally, in this line:

printf ("%c ", arr1[i]);

Use %d as format specifier instead: you want to print a number, not a character.

score 0 · Answer 4 · answered May 13 '13 at 12:27

#include <stdio.h>

int main(void){
    int arr1[10] = {0};
    char table[] = "0123456789";
    int c;

    while ((c = getchar()) != '\n'){
        for (int i = 0; i <= 9; i++){
            if (c == table[i])//or table[i] --> "0123456789"[i]
                arr1[i] += 1;
        }
    }
    for (int i = 0; i <= 9; i++)
        printf ("%d ", arr1[i]);
}

Using getchar() to store a value to compare with an integer

4 Answers4