what does putchar('0' + num); do?

Question

I am trying to understand how the putchar('0' + r); works. Below, the function takes an integer and transform it to binary.

void to_binary(unsigned long n)
{
   int r;
   r = n % 2;
   if (n >= 2)
      to_binary(n / 2);
   putchar('0' + r);
}

I google the definition of putchar but I didn't find this. To test it, I added a printf to see the value of the r:

void to_binary(unsigned long n)
{
   int r;
   r = n % 2;
   if (n >= 2)
      to_binary(n / 2);
   printf("r = %d and putchar printed ", r);
   putchar('0' + r);
   printf("\n");
}

and I run it (typed 5) and got this output:

r = 1 and putchar printed 1
r = 0 and putchar printed 0
r = 1 and putchar printed 1

So I suppose that the putchar('0' + r); prints 0 if r=0, else prints 1 if r=1, or something else happens?

Answer 1

In C '0' + digit is a cheap way of converting a single-digit integer into its character representation, like ASCII or EBCDIC. For example if you use ASCII think of it as adding 0x30 ('0') to a digit.

The one assumption is that the character encoding has a contiguous area for digits - which holds for both ASCII and EBCDIC.

---

As pointed out in the comments this property is required by both the C++ and C standards. The C standard says:

5.2.1 - 3

In both the source and execution basic character sets, the value of each character after 0 in the above list of decimal digits shall be one greater than the value of the previous.