Contributor: SWAG SUPPORT TEAM { >Also, how would I simply read each bit? } { Test if a bit is set. } Function IsBitSet(Var INByte : Byte; Bit2Test : Byte) : Boolean; begin if (Bit2Test in [0..7]) then IsBitSet := ((INByte and (1 shl Bit2Test)) <> 0) else Writeln('ERROR! Bit to check is out of range!'); end; { IsBitSet. } { >How on earth can I manipulate an individual bit? ...One method is to use the bit-operators: AND, OR, XOR, NOT } { Manipulate an individual BIT within a single Byte. } Procedure SetBit(Bit2Change : Byte; TurnOn : Boolean; Var INByte : Byte); begin if Bit2Change in [0..7] then begin if TurnOn then INByte := INByte or (1 shl Bit2Change) else INByte := INByte and NOT(1 shl Bit2Change); end; end; { SetBit. } { >...but I'm not sure exactly what the shifting is doing. } { Check if the bit is to be turned on or off. } If TurnOn then { SHL 1 (which has a bit map of 0000 0001) to the bit position we want to turn-on. ie: 1 SHL 4 = bit-map of 0001 0000 ...Then use a "logical OR" to set this bit. ie: Decimal: 2 or 16 = 18 Binary : 0000 0010 or 0001 0000 = 0001 0010 } INByte := INByte or (1 shl Bit2Change) else { Else turn-off bit. SHL 1 (which has a bit map of 0000 0001) to the bit position we want to turn-off. ie: 1 SHL 4 = bit-map of 0001 0000 ...Then use a "logical NOT" to flip all the bits. ie: Decimal: not ( 16 ) = 239 Binary : not (0001 0000) = (1110 1111) ...Than use a "logical AND" to turn-off the bit. ie: Decimal: 255 and 239 = 239 Binary : 1111 1111 and 1110 1111 = 1110 1111 } INByte := INByte and NOT(1 shl Bit2Change); { >Also, how can you assign a Byte (InByte) a Boolean value (OR/AND/NOT) or / xor / and / not are "logical" bit operators, that can be use on "scalar" Types. (They also Function in the same manner For "Boolean" logic.) >If I have, say 16 bits in one Byte, the interrupt list says that for >instance the BIOS calls (INT 11), AX is returned With the values. It >says that the bits from 9-11 tell how many serial portss there are. >How do I read 3 bits? To modify the two routines I posted wo work With 16 bit Variables, you'll need to change: INByte : Byte; ---> INWord : Word; ...Also: in [0..7] ---> in [0..15] ...If you don't want to use the IsBitSet Function listed above (modified to accept 16-bit Word values) you could do the following to check if bits 9, 10, 11 are set in a 16-bit value: The following is the correct code For reading bits 9, 10, 11 of the 16-bit Variable "AX_Value" : Port_Count := ((AX_Value and $E00) SHR 9); NOTE: Bit-map For $E00 = 0000 1110 0000 0000 ...If you've got a copy of Tom Swan's "Mastering Turbo Pascal", check the section on "logical operators". { >Var Regs : Registers; >begin > Intr($11,Regs); > Writeln(Regs.AX); >end. >How do I manipulate that to read each bit (or multiple bits like >the number of serial ports installed (bits 9-11) ? } Uses Dos; Var Port_Count : Byte; Regs : Registers; begin Intr($11, Regs); Port_Count := ((Regs.AX and $E00) SHR 9); Writeln('Number of serial-ports = ', Port_Count) end. { NOTE: The hex value of $E00 is equivalent to a 16-bit value with only bits 9, 10, 11 set to a binary 1. The SHR 9 shifts the top Byte of the 16-bit value, to the lower Byte position. } { >Is $E00 the same as $0E00 (ie, can you just omit leading zeros)? Yeah, it's up to you if you want to use the leading zeros or not. The SHR 9 comes in because once the value has been "AND'd" with $E00, the 3 bits (9, 10, 11) must be placed at bit positions: 0, 1, 2 ...to correctly read their value. For example, say bits 9 and 11 were set, but not bit 10. If we "AND" this With $E00, the result is $A00. 1011 1010 0111 1110 and 0000 1110 0000 0000 = 0000 1010 0000 0000 ^ ^ (bits 9,11 are set) and ( $E00 ) = $A00 ...Taking the result of $A00, and shifting it right 9 bit positions $A00 SHR 9 = 5 0000 1010 0000 0000 SHR 9 = 0000 0000 0000 0101 ...Which evalutates to 5. (ie: 5 serial ports) } { Get Equipment Bit-Map --------------------- AH AL 76543210 76543210 AX = ppxgrrrx ffvvmmci ... ... rrr = # of RS232 ports installed ... ... (* reports the number of RS232 ports installed *) Function NumRS232 : Byte; Var Regs : Registers; (* Uses Dos *) begin Intr($11,Regs); NumRS232 := (AH and $0E) shr 1; end; ...When you call Int $11, it will return the number of RS232 ports installed in bits 1-3 in register AH. For example if AH = 01001110 , you can mask out the bits you *don't* want by using AND, like this: 01001110 <--- AH and 00001110 <---- mask $0E ÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 00001110 <---- after masking Then shift the bits to the right With SHR, 00001110 <---- after masking SHR 1 <---- shift-right one bit position ÄÄÄÄÄÄÄÄÄÄÄÄÄ 00000111 <---- result you want } { -> How do I know to use $4 For the third bit? Suppose I want to read -> the fifth bit. Do I simply use b := b or $6? Binary is a number system just like decimal. Let me explain. First, consider the number "123" in decimal. What this means, literally, is 1*(10^2) + 2*(10^1) + 3*(10^0), which is 100 + 20 + 3. Binary works just the same, however instead of a 10, a 2 is used as the base. So the number "1011" means 1*(2^3) + 0*(2^2) + 1*(2^1) + 1*(2^0), or 8+0+2+1, or 11. This should make it clear why if you wish to set the nth bit to True, you simply use a number equal to 2^(n-1). (The -1 is there because you probably count from 1, whereas the powers of two, as you may note, start at 0.) -> b or (1 SHL 2) Would mean that b := 1 (True) if b is already equal to -> one (1) and/OR the bit two (2) to the left is one (1) ??? Aha. You are not familiar With bitwise or operations. When one attempts to or two non-Boolean values (Integers), instead of doing a logical or as you are familiar with, each individual BIT is or'd. I.E. imagine a Variables A and B had the following values: a := 1100 (binary); b := 1010 (binary); then, a or b would be equal to 1110 (binary); Notice that each bit of a has been or'd With the corresponding bit of b? The same goes For and. Here's an example. a := 1100 (binary); b := 1010 (binary); a and b would be equal to 1000; I hope this clears up the confusion. And just to be sure, I'm going to briefly show a SHL and SHR operation to make sure you know. Consider the number a := 10100 (binary); This being the number, A SHL 2 would be equal to 1010000 (binary) -- notice that it has been "shifted to the left" by 2 bits. A SHR 1 would be 1010 (binary), which is a shifted to the right by 2 bits. }