Return a single value using AND OR...Help.

[typosoft]

I have a function that detects if a key is pressed and it returns a value. How do I make the function return that UP and LEFT were pressed? I've seen in other people's code the use of AND OR or something, so could someone help me. DOWN = 1 RIGHT = 2 UP = 3 LEFT = 4.

[toonski84]

you need to use separate digits of any system.

for instance, if you wanted to use arabic numerals:

1010 - one thousand and ten.

but for total efficiency, use binary:

1010 - ten.

to read a binary digit:

is_it_on = x and (2^digit - 1) - qbasic
is_it_on = x & (exp(2, digit) - 1); - c/c++ [i think]

[typosoft]

I made the function return a string and then I parse the string for commands that it gave. But it will still only take one command at a time. Here's code:

Code:

std::string s = getInput();
    if (!s.find("RIGHT")) {
        xv+= ACCEL;
    }
    if (!s.find("UP")) {
        y--;
        //yv-= ((ACCEL * 255) * YFRICTION);
    }

Code:

std::string cBubble::getInput()
{
    std::string s;
    if (key[KEY_ESC]) return "ESC";
    if (key[KEY_RIGHT]) s+="RIGHT";
    if (key[KEY_LEFT]) s+="LEFT";
    if (key[KEY_DOWN]) s+="DOWN";
    if (key[KEY_UP]) s+="UP";
    return s;
}

[na_th_an]

I would do something like this:

Code:

UP = bit 1
DOWN = bit 2
LEFT = bit 3
RIGHT = bit 4

Then I just...

Code:

pressed = 0;

if (key[KEY_UP])
   pressed = pressed | 1;
else
   if (key[KEY_DOWN])
      pressed = pressed | 2;

if (key[KEY_LEFT])
   pressed = pressed | 4;
else
   if (key[KEY_RIGHT])
      pressed = pressed | 8;

return pressed;

Then, to check if left, for example:

Code:

if ( (pressed & 4) == 4 )
   // left was pressed.

if ( (pressed & 5) == 5)
   // left and up were pressed ( 4 + 1 = 5, bits 1 _AND_ 3).

Bit #n's power of two is calculated this way (n ranges from 1 to...):

Code:

power_of_two = 1 << (n-1)

[typosoft]

Awesome. That's exactly what I wanted. I got it running, but didn't use that last part. What is that and what does it do?

Bit #n's power of two is calculated this way (n ranges from 1 to...):

Code:
power_of_two = 1 << (n-1)

and if I wanted to add in other options to it, I just add in another number 2x the size of the last one?

[na_th_an]

It calculates a power of two without multiplying.

When you are dealing with bits 1, 2, 3 and stuff you will remember the values (1,2,4,8...) but which number will you use to AND/OR to modify bit #27?

Code:

mask = 1 << (27-1);

This is used if the bit # is a variable. To print a number (unsigned short 16 bits integer) in binary:

Code:

for (i=16;i>=1;i++)
{
   mask = 1<<(i-1);
   printf("%i", (number & mask) / mask);
}

[relsoft]

<< =Shift Left

>> = shift Right

[Ninkazu]

<< =Shift Left

>> = shift Right

isn't >>> the right shift that doesn't wrap?

[Hard Rock]

erm, afaik >> doesnt wrap. It just kicks off the extra bits.

[relsoft]

<< =Shift Left

>> = shift Right

isn't >>> the right shift that doesn't wrap?

Shr and Shl does not wrap

there is a Rotate Bits function in ASM. Dunno the C flavor though.