Question

I’m trying to understand the following function which decides whether a bit is on:

int isBitISet( char ch, int i )
{
   char mask = 1 << i ;
   return mask & ch ;
}

First, why do I get a char? for ch=abcdefgh and i=5 the function suppose to return the fifth bit from the right (?) , d. so mask=00000001<<5=00100000, and 00100000 & abcdefgh = 00c00000.

Can you please explain me how come we get char and we can do all these shifts without any casting? how come we didn’t get the fifth bit and why the returned value is really the Indication whether the bit is on or not?

Edit: the ‘abcdefg’ are just a symbols for the bits, I didn’t mean to represent a string in a char type.

I used to think of a char as 'a' and not as an actual 8 bits, so probably this is the answer to my first question.

Answer 1

It won’t give you the fifth bit. Binary numbers start at 2⁰, so the first bit is actually indexed with 0, not with 1. It will give return you sixth bit instead.

Examples:

ch & (1 << 0); // first bit
ch & (1 << 1); // second bit
ch & ((1 << 3) | (1 << 2)); // third and fourth bit.

Also, a char is only an interpretation of a number. On most machines it has a size of 8 bit, which you can either interpret as a unsigned value (0 to 255) or signed value (-128 to 127). So basically it’s an integer with a very limited range, thus you can apply bit shifting without casting.

Also, your function will return an integer value that equals zero if and only if the given bit isn’t set. Otherwise it’s a non-zero value.