Hi,
I have a question concerning the saving of the keys in bits of a skill. It worked till I wanted to read the 7th bit. The var check3 showed 0. This is how I wrote it:

my.skill1 = 0;
my.skill1 |= 1;
my.skill1 |= 1<<1;
my.skill1 |= 1<<2;
my.skill1 |= 1<<3;
my.skill1 |= 1<<4;
my.skill1 |= 1<<5;
my.skill1 |= 1<<6;
my.skill1 |= 1<<7;
check2 = (my.skill1&0x00000001);
check2 = (my.skill1&0x00000002)>>1;
check2 = (my.skill1&0x00000004)>>2;
check2 = (my.skill1&0x00000008)>>3;
check2 = (my.skill1&0x000000016)>>4;
check2 = (my.skill1&0x000000032)>>5;
check2 = (my.skill1&0x000000064)>>6;
check3 = (my.skill1&0x0000000128)>>7;
Thanks for any help

You are mixing hexadecimal with decimal.

Hex:
0x01
0x02
...
0x09
0x0A
...
0x0F
0x10

So you could write it this way (decimal):
(my.skill1 & 1)
(my.skill1 & 2) >> 1
...
(my.skill1 & 128) >> 7

Is there a site where this is explained? I just pasted it from your Anet-manuel in the tipps & tricks section and thought it works analog till the 32nd bit. I didn`t even understand what the "0x000..." actually stands for.
thanks for the help

Hexadecimal is explained here: http://en.wikipedia.org/wiki/Hexadecimal

Maybe this helps you understand how using flags (= using single bits of a variable) works: http://www.rohitab.com/discuss/index.php?showtopic=25162

Hm sorry could you show me an example how to write/read the last bit of a skill? I still quite understand how it works...
thanks a lot

A skill is a var (= float). Which means it consists of 4 byte. Each byte has 8 bit.

--> 4 Byte = 32 bit

If you want to set the last bit of a skill you have to do a OR operation with
1000 0000 0000 0000 0000 0000 0000 0000 (binary!).
--> As you see the 32. bit is 1
Because we can't write binary numbers in lite-c, we have to calculate the according hex or decimal number:
0x80000000 (hex)
2147483648 (Dec)

my.skill[1] = my.skill[1] | 0x80000000; //sets the 32. bit

For clearing a bit, you have to make an AND operation with 0111 1111 1111 1111 1111 1111 1111 1111 (binary) = 0x7FFFFFFF (hex)

my.skill[1] = my.skill[1] & 0x7FFFFFFF; //resets the 32. bit


This may sound complicated but if you look into binary operations (AND/OR) and how binary numbers and hexadecimal works it is pretty easy

but if i set the last bit of skill1 shouldn`t it have the value 2147483648 afterwards if i set skill1 to zero before? Because only the value zero is displayed instead of 2147483648.

But thanks for the explanation it helped a lot
If i want to read out the value of the last bit should it look something like this? :

right_key = (my.skill1 & 0x80000000)>>31

and thanks for the quick answer

Exactly!

The problem is, that skills are floats => 10000...(binary) != 2147483648

If you will do this with an int instead of a float, you will see that it is 2147483648 after the OR operation.