Why I should subtract '0' so that I can do int_arr[5] = char_string[5]?

Question

So I was assigned to store two 50 digit integers in c language and do math equations using them. the problem for me was to store the input digit by digit in an array. I figured that I can store the input in a char string like this:

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

int main ()
{
    char string_test [50];
    scanf("%s", string_test);
    return 0;
}

But I couldn't use it as a number because it was stored as char and I couldn't copy them digit by digit into another array which was defined as int

after a whole day of searching, I found out that I need to copy my string one by one like this:

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

int main ()
{
    char string_test [50];
    scanf("%s", string_test);

    int arr[50];

    for (int i = 0; i < 50; i++)
    {
        arr[i] = string_test[i] - '0';
    }
    return 0;
}

Now my question is why do I need to subtract '0' to get a suitable result?

Aside: `char string_test [50];` 1 too small to store a 50 digit _string_. Use `char string_test [50+1]; scanf("%50s", string_test);` or the like. — chux - Reinstate Monica, Oct 20 '21 at 13:02
You might look up the difference between a numerical value `5` and a ASCII representation for character `'5'`. The latter has numerical value `0x35` or `53` which is rather useless for doing math calculations — Gerhardh, Oct 20 '21 at 13:04
The answer to your questions and the explanation to your problems is found in the chapter called "strings" of your average C programming book. It should take much less time to study than a whole day of searching. — Lundin, Oct 20 '21 at 13:05
The digit characters `'0'` to `'9'` are guaranteed to have consecutive character codes, so subtracting the character code of the `'0'` character from any digit character will convert it to the corresponding digit value. — Ian Abbott, Oct 20 '21 at 13:22

ryyker · Accepted Answer · 2021-10-20T14:22:40.623

The ASCII values for digits 0 - 9 are:

Digit          0   1   2   3   4   5   6   7   8   9

ASCII value   48  49  50  51  52  53  54  55  56  57

So if you have a string representation of an integer, say

char int_str[] = "123456";

and need to convert each char to its numeric value, subtracting the value for '0' (48) from each will will result in the values

int_str[0] == '1' ==>  '1' - '0' ==> 42 - 41 == 1
int_str[1] == '2' ==>  '2' - '0' ==> 43 - 41 == 2
int_str[2] == '3' ==>  '3' - '0' ==> 44 - 41 == 3
int_str[3] == '4' ==>  '4' - '0' ==> 45 - 41 == 4
int_str[4] == '5' ==>  '5' - '0' ==> 46 - 41 == 5
int_str[5] == '6' ==>  '6' - '0' ==> 47 - 41 == 6

To get digits 1 2 3 4 5 6 into the integer 123456 requires additional steps:

This example uses the same conversions encapsulated into a function to convert discrete char digit to int digit values, then assimilate each discrete int digit value into the composite integer value:

int main(void)
{
    char str[] = "123456";
    int int_num = str2int(str);

    return 0;   
}

 int str2int(char *str)
 {
    int sum=0;

    while(*str != '\0')
    {    //qualify string
         if(*str < '0' || *str > '9')
         {
             printf("Unable to convert it into integer.\n");
             return 0;
         }
         else
         {   //assimilate digits into integer 
             sum = sum*10 + (*str - '0');
             str++;
         }
    }
    return sum;
}

score 0 · Answer 2 · answered Oct 20 '21 at 13:25

A char like '0' is just an ASCII representation of a 1 byte number. The numeric representations of these characters can be found in the manual (man ascii). Here you will see that '0' actually represents number 48 or 0x30 and that ASCII '0' to '9' are consecutive.

To convert a numeric char value to its integer counterpart, it is necessary to subtract this value of 48 or '0'.

I hope this clears things up. For further info, look into char arithmetic in C.

score 0 · Answer 3 · edited Oct 20 '21 at 14:21

This is actually related to the ASCII code representation of numbers; when you input a character, '0', it is stored in memory as value 48 (in decimal) and also any other number or character.

You will find that the ASCII code value for '1' is 49 so, if we applied the operation, int x = '1' - '0';, we are storing the decimal value 1 in x and dealing with it as a number, not a character any more.

You can search more about ASCII codes; it's a very useful topic for any programmer.

Why I should subtract '0' so that I can do int_arr[5] = char_string[5]?

3 Answers3