AND
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The logical AND numerical operator compares two values in respect of their bits. If both bits at a certain position in both values are set, then that bit position is set in the result.
Syntax
- result = firstvalue AND secondvalue
Description
- AND compares the bits of the firstvalue against the bits of the secondvalue, the result is stored in the result variable.
- If both bits are on (1) then the result is on (1).
- All other conditions return 0 (bit is off).
- AND is often used to see if a bit is on by comparing a value to an exponent of 2.
- Can turn off a bit by subtracting the bit on value from 255 and using that value to AND a byte value.
Examples
Example 1: Template:WhiteStart
101
AND
011
-----
001
- The 101 bit pattern equals 5 and the 011 bit pattern equals 3, it returns the bit pattern 001 which equals 1. Only the Least Significant Bits (LSB) match. So decimal values 5 AND 3 = 1.
Example 2:
Template:WhiteStart
11111011
AND
11101111
----------
11101011
- Both bits have to be set for the resulting bit to be set. You can use the AND operator to get one byte of a two byte integer this way:
- firstbyte = twobyteint AND 255
- Since 255 is 11111111 in binary, it will represent the first byte completely when compared with AND.
- To find the second (HI) byte's decimal value of two byte INTEGERs use: secondbyte = twobyteint \ 256
Example 3: Finding the binary bits on in an INTEGER value.
DO INPUT "Enter Integer value from -32768 to 32767 (Enter quits): ", INTvalue& IF INTvalue& < -32768 OR INTvalue& > 32767 OR INTval& = 0 THEN EXIT DO FOR exponent = 15 TO 0 STEP -1 IF (INTvalue& AND 2 ^ exponent) THEN PRINT "1"; ELSE PRINT "0"; NEXT PRINT " " LOOP UNTIL INTvalue& = 0 'zero entry quits |
- Example output for 6055.
0001011110100111 |
- Note: The value of 32767 sets 15 bits. -1 sets all 16 bits. Negative values will all have the highest bit set. Use LONG variables for input values to prevent overflow errors.
See also