Programming joysticks

[deltacoder1020]

here they are in binary, decimal, and hex forms...

Neutral: 001011xx 44 0x2C
Up Left: 111111xx 252 0xFC
Up: 111111xx 252 0xFC
Up Right: 111111xx 252 0xFC
Left: 011000xx 96 0x60
Right: 110010xx 200 0xCB
Down Left: 101001xx 164 0xA4
Down: 100101xx 148 0x94
Down Right: 110101xx 212 0xD4

Here's a diagram that gives the full range of values and the value to test with AND is bold.

252-255*252-255*252-255
*96-99***44-47**200-203
164-167*148-151*212-215

[Robert Hafner]

Because the last bits are unreliable (from what i've read), how will this affect the program? Should I stick with a general area? example:

if (p1_wheel >= 42 && p1_wheel <= 46)
    // I am neutral
if (p1_wheel >= 250 && p1_wheel <= 254)
     // I am upperleft;

[deltacoder1020]

your previous code with the bitwise AND (&) should work fine - just use the hex or decimal codes I posted above (because you AND it with the original constant, the lower bits don't matter, and thus it will be true for the entire range).

[Robert Hafner]

at the risk of sounding dumb, how does the AND thing work, and what do I need to do with it?

[deltacoder1020]

if( (p1_wheel & 0xC0) == 0xC0 )

"&" is the "bitwise AND" operator. that means that it looks at individual bits of a value, and does an AND operation. AND is defined as follows: true if both inputs are true, otherwise false. a logic table:

0, 0 = 0
1, 0 = 0
0, 1 = 0
1, 1 = 1

so as you can see, it will only return 1's in the places where both numbers have ones. thus, it will only return all of the bits of one value if the same bits are 1's in the other value. the if statement essentially checks to see that all of the bits that are 1 in 0xC0 are also 1 in p1_wheel. replace 0xC0 with whatever "bitmask" (set of bits) you want to test as being 1.