merging of two character constants in c

Question

I want to merge two single character constants into one like '1'+'2'='12'.So that I can add an integer value to this single character. I have tried a lot of things like used their ASCII values.Also the substring functions doesn't seem to work in this context.Help!

Answer 1

If you want to compute the number from 2 digit characters, the formula is very simple:

char c1 = '1';
char c2 = '2';
int value = (c1 - '0') * 10 + (c2 - '0');  // value is 12

c1 - '0' evaluates to the number represented by the digit c1. It works because the digits are guaranteed to be consecutive from '0' to '9' in the execution character set.

Answer 2

There are more 'beautiful' ways to do it, but this ought to be enough for a first attempt.

int value = 0;

if (firstcharacter == '1') value = 10;
if (firstcharacter == '2') value = 20;
/* ... */
if (firstcharacter == '9') value = 90;

if (secondcharacter == '1') value += 1;
/* ... */
if (secondcharacter == '9') value += 9;

printf("value + 42 is %d.\n", value + 42);

Answer 3

If the goal is to combine two chars which are digits to form an integer value that can be used in arithmetic expressions, then a compound literal can be used to form a string from the chars, and strtol() can be used to convert the string to an integer value. Here is an example that uses the function char_digits_int() to convert a pair of char inputs into an int value. If either of the arguments is not a digit, -1 is returned:

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

int char_digits_int(char x, char y);

int main(void)
{
    char a = '1';
    char b = '2';

    printf("%c and %c form: %d\n", a, b, char_digits_int(a, b));
    printf("The sum of %c%c and 1 is: %d\n",
           a, b, char_digits_int(a, b) + 1);

    int value = char_digits_int('0', 'x');
    if (value == -1) {
        fprintf(stderr, "Non-digit input\n");
    } else {
        printf("Value: %d\n", value);
    }

    return 0;
}

/* Returns -1 if x or y is not a digit */
int char_digits_int(char x, char y)
{
    int val = -1;

    if (isdigit((unsigned char) x) && isdigit((unsigned char) y)) {
        val = strtol((char []) { x, y, '\0' }, NULL, 10);
    }
    return val;
}

Program output:

1 and 2 form: 12
The sum of 12 and 1 is: 13
Non-digit input

Answer 4

If your goal is to combine the character constants '1' and '2' to get the value of the multi-character character constant '12', the answer is There is no portable way to do this and you should probably not write code that uses these things anyway.

Look at this answer for a historical perspective on this oddity: https://stackoverflow.com/a/45576458/4593267

It you are still curious (good for you), here is a test case:

#include <limits.h>
#include <stdio.h>

int main(void) {
    if ((('1' << CHAR_BIT) | '2') == '12')
        printf("(('1' << CHAR_BIT) | '2') == '12'\n");
    else
    if ((('2' << CHAR_BIT) | '1') == '12')
        printf("(('2' << CHAR_BIT) | '1') == '12'\n");
    else
        printf("No simple way to get '12' from '1' and '2'\n");
    return 0;
}

On my system, I get these warnings:

multichar.c:5:38: warning: multi-character character constant [-Wmultichar]
    if ((('1' << CHAR_BIT) | '2') == '12')
                                     ^
multichar.c:8:38: warning: multi-character character constant [-Wmultichar]
    if ((('2' << CHAR_BIT) | '1') == '12')
                                     ^
multichar.c:8:11: warning: code will never be executed [-Wunreachable-code]
    if ((('2' << CHAR_BIT) | '1') == '12')
          ^~~
multichar.c:9:9: warning: code will never be executed [-Wunreachable-code]
        printf("(('2' << CHAR_BIT) | '1') == '12'\n");
        ^~~~~~
4 warnings generated.

And this output:

(('1' << CHAR_BIT) | '2') == '12'

But other systems, notably big endian ones, may produce a different output.