Converting integer to string in C

Question

int x =5;
char y = x + '0';

Apparently this bit of code converts an integer to a string. But I don't understand how it works behind the scenes. I'm a newbie. Could someone explain how adding a string 0 with an int converts it into a string?

Answer 1

In

char y = x + '0';

put the character code (ASCII or whatever character code your system implements) of '0' and add 5 to it then again convert that integer to an ASCII(/UTF-8 or any other character encoding) character that's what your output(5) will be.

Behind the scenes this is what's happening:

y = 5 + 48  ;for ASCII/UTF-8 character code of '0' is 48
y = 53

and ASCII/UTF-8 char for 53 is '5'.

It does not convert a string to an integer. It just gives you the equivalent character. We can convert an int(0-127) to a char or vice-versa, output will be an ASCII/UTF-8 value. char are stored as short int in memory. i.e. they are 8-bit int values.

Answer 2

You have a bit of misunderstanding here. The variable y isn't string, it's a char. string is an array of chars. What you are actually doing here is:

char y = x + 48; //48 is the ASCII encoding of 0

This is somewhat like:

char y = 50;
printf("%c\n", y); //prints 2 as a char

This is because chars are encoded with a specific value. Eg. 'a' is 97. 'b' is 98 and so on. The same way '0''s code is 48.

 char y = x + '0'

means is adding the encoded value of 0(ie 48) to x. And because 0 is the first number, the other number's code will be right next to it. This is what I mean:

48 -> This is 0
49 -> This is 1
50 -> This is 2
51 -> This is 3
52 -> This is 4
53... and so on

You may notice that in order to get eg. 1, we need to add one to zero. 48 + 1 = 49 and 49 is the encoding for 1. This is true for all numbers.

Note: I've used ASCII encoding for explaining but there are others and it should work on most on weird encodings.

Answer 3

'0' is an int, not a string. String literals have double quotes (e.g. "Hi!"). Character literals (which are integers) have single quotes (e.g. 'A').

x + '0' evaluates to int since both the operands are int.

The integral value of a character literal is its ASCII value : https://en.wikipedia.org/wiki/ASCII#Character_set
From the ascii table, the integer value of '0' is 48.

Now, coming to the expression in the question:
x is 5, '0' is 48.
Therefore y = 48+5 (i.e 53).

When you print y as a character using printf("%c", y), the character whose ASCII value is 53 is printed. i.e. '5'

Why adding '0' to a single digit integer seems to convert it into an integer is because '0' to '9' are one after another in the ASCII table. This means '9' (57) is exactly 9 spaces after '0' (48).

This is why adding double digit integers to '0' will not work. E.g. '0' + 10 will be ':' (ASCII value 58).

To properly convert larger strings to integers, use the atoi function (https://man7.org/linux/man-pages/man3/atoi.3.html). e.g.

int a = atoi("1023"); // a is an integer with the value 1023