Bitwise AND usage problem understanding

[alMarshy]

Im reading a c++ beginners book and there is a chapter about & (bitwise AND). It is using this to convert a lowercase character to its uppercase equivalent. I cant for the life of me figure out why the value 223 is used after the bitwise AND

for(int i=0;i<19;1++)
char ch='a'+i;

cout<<ch<<endl;

//convert to uppercase using bitwise AND
ch=ch & 223;
cout<<"Uppercase :"<<ch<<endl;

The book says that the bitwise AND can be used to turn off the 6th bit in order to make a character its uppercase equivalent. So why not use :

ch=ch & 32;

I guess I am lost as to the value that 223 is or something? Can anybody explain it to me?

thanks in advance.

[guestgulkan]

An upper case letter is 32 decimal (20 hex) less than it's equivalent lower case partner.
So take the lower case a - in decimal 97, in hex 61 in binary 0110 0000
to change to upper case we want to clear only the bit 5 leaving all the other bit's untouched.
When using AND logic, to leave a bit untouched, we AND with 1, to clear it we AND with 0.

so we will need to AND with 1101 1111 which is 223 decimal.

[Disch]

A better way to write it would be like this:

ch &= ~0x20;

The ~ operator makes it easier to see that you're turning a specific bit off.

Hex makes it easier to see which bit you're turning off.

[m4ster r0shi]

ch=ch & 32; turns off ALL bits except bit number 6.

ch=ch & ~32; is equivalent to ch=ch & 223; and clears the 6th bit.

ch=ch ^ 32; toggles the 6th bit. You can use it to toggle the case.

[alMarshy]

Thanks a million everybody that has cleared it up for me

[guestgulkan]

You will notice though - that you cannot use AND to set a bit to 1

[m4ster r0shi]

That's true. To set a bit you'll have to use the bitwise OR operator.
This, for example, turns on the 6th bit: ch=ch | 32;