how to add char type integer in c

Question

This is the sample code of my program, in which i've to add two string type integer (ex: "23568" and "23674"). So, i was trying with single char addition.

char first ='2';
char second ='1';

i was trying like this..

i=((int)first)+((int)second);
printf("%d",i);

and i'm getting output 99, because, it's adding the ASCII value of both. Anyone please suggest me, what should be the approach to add the char type number in C.

Answer 1

Since your example has two single chars being added together, you can be confident knowing two things

1. The total will never be more than 18.
2. You can avoid any conversions via library calls entirely. The standard requires that '0' through '9' be sequential (in fact it is the only character sequence that is mandated by the standard).

Therefore;

char a = '2';
char b = '3';

int i = (int)(a-'0') + (int)(b-'0');

will always work. Even in EBCDIC (and if you don't know what that is, consider yourself lucky).

If your intention is to actually add two numbers of multiple digits each currently in string form ("12345", "54321") then strtol() is your best alternative.

Answer 2

i=(first-'0')+(second-'0');

No need for casting char to int.

Answer 3

if you want to add the number reprensations of the characters, I would use "(first - '0') + (second - '0');"

Answer 4

The question seemed interesting, I though it would be easier than it is, adding "String numbers" is a little bit tricky (even more with the ugly approach I used).

This code will add two strings of any length, they doesn't need to be of the same length as the adding begins from the back. Your provide both strings, a buffer of enough length and you ensure the strings only contains digits:

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

char * add_string_numbers(char * first, char * second, char * dest, int dest_len)
{
    char * res = dest + dest_len - 1;
    *res = 0;

    if ( ! *first && ! *second )
    {
        puts("Those numbers are less than nothing");
        return 0;
    }

    int first_len = strlen(first);
    int second_len = strlen(second);

    if ( ((first_len+2) > dest_len) || ((second_len+2) > dest_len) )
    {
        puts("Possibly not enough space on destination buffer");
        return 0;
    }

    char *first_back = first+first_len;
    char *second_back = second+second_len;

    char sum;
    char carry = 0;

    while ( (first_back > first) || (second_back > second) )
    {
        sum = ((first_back  > first)  ? *(--first_back) : '0')
            + ((second_back > second) ? *(--second_back) : '0')
            + carry - '0';

        carry = sum > '9';
        if ( carry )
        {
            sum -= 10;
        }

        if ( sum > '9' )
        {
            sum = '0';
            carry = 1;
        }

        *(--res) = sum;
    }

    if ( carry )
    {
        *(--res) = '1';
    }

    return res;
}

int main(int argc, char** argv)
{
    char * a =              "555555555555555555555555555555555555555555555555555555555555555";
    char * b = "9999999999999666666666666666666666666666666666666666666666666666666666666666";
    char r[100] = {0};

    char * res = add_string_numbers(a,b,r,sizeof(r));

    printf("%s + %s = %s", a, b, res);

    return (0);
}

Answer 5

Well... you are already adding char types, as you noted that's 49₁₀ and 50₁₀ which should give you 99₁₀

If you're asking how to add the reperserented value of two characters i.e. '1' + '2' == 3 you can subtract the base '0':

printf("%d",('2'-'0') + ('1'-'0')); 

This gives 3 as an int because:

'0' = ASCII 48<sub>10</sub>
'1' = ASCII 49<sub>10</sub>
'2' = ASCII 50<sub>10</sub>

So you're doing:

printf("%d",(50-48) + (49-48));

If you want to do a longer number, you can use atoi(), but you have to use strings at that point:

int * first = "123";
int * second = "100";
printf("%d", atoi(first) + atoi(second));

>> 223

Answer 6

In fact, you don't need to even type cast the chars for doing this with a single char:

 #include <stdlib.h>
 #include <stdio.h>
 int main(int argc, char* argv[]) {
     char f1 = '9';
     char f2 = '7';
     int v = (f1 - '0') - (f2 - '0');
     printf("%d\n", v);

     return 0;
 }

Will print 2

But beware, it won't work for hexadecimal chars

Answer 7

This will add the corresponding characters of any two given number strings using the ASCII codes.

• Given two number strings 'a' and 'b', we can compute the sum of a and b using their ASCII values without type casting or trying to convert them to int data type before addition.

Let

char *a = "13784", *b = "94325";
int max_len, carry = 0, i, j; /*( Note: max_len is the length of the longest string)*/
char sum, *result;

Adding corresponding digits in a and b.

    sum = a[i] + (b[i] - 48) + carry; /*(Because 0 starts from 48 in ASCII) */

    if (sum >= 57)
        result[max_len - j] = sum - 10;
        carry = 1;
    else
        result[max_len - j] = sum;
        carry = 0;



 /* where (0 < i <= max_len and 0 <= j <= max_len) */

NOTE: The above solution only takes account of single character addition starting from the right and moving leftward.

Answer 8

if you want to scan number by number, simple atoi function will do it

Answer 9

you can use atoi() function

#include <stdio.h>
#include <stdlib.h>
void main(){
    char f[] = {"1"};
    char s[] = {"2"};
    int i, k;
    i = atoi(f);
    k = atoi(s);
    printf("%d", i + k);
    getchar();
}

Hope I answered you question