Fifth Steps =========== In the last tutorial we looked at subroutines. Now we are going to move on to binary operations. Before doing so lets cover some of the terminology: Terminology ----------- Binary In computer junky terminology binary is refering to the base 2 number system which consists of the digits 0 and 1. Bit A bit is a binary digit, ie 0 or 1. All information is stored in a computer as sets of bits. Bit Set A bit is said to be set if its value is 1. Bit Clear A bit is said to be clear if its value is 0. Byte A byte is a set of eight bits. We often wish to refer to an individual bit in a byte and for this reason a each bit in a byte has a number depending on its position. The bits are numberd from right to left starting at 0. So bit 0 refers to the right most bit of a byte and bit 7 refers to the left most bit. Take for example the byte 1011 0110 : bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 1 0 1 1 0 1 1 0 Word A word is a set of 16 bits. Obviously then a word is equivalent to two bytes. The same bit numbering system is adopted as for a byte, the bits being numberd 0 to 15 from right to left. Long Word A long word is a set of 32 bits, or 2 words, or 4 bytes. The bits are numberd 0 to 31 from right to left. MSB Most Signifigant Bit. This is the bit with the highest reference number. So in a byte, the MSB will be bit 7, in a word the MSB will be bit 15 and in a long word the MSB will be bit 31. LSB Least Signifigant Bit. This is always bit 0 wether refering to a byte, word or long word. Sign Bit If dealing with signed numbers, the MSB is sometimes refered to as the Sign Bit. As you now know, the internal registers are all 32 bits long ( ie they can hold long word values ). With most 68000 instructions you can specify the size of data you can operate on as byte ( .b ) word ( .w ) or long word (.l ). If you specify .l then bits 0 to 31 are effected. If you specify .w the bits 0 to 15 are effected and if you specify .b then bits 0 to 7 are effected. This was demonstrated earlier in the series, but to clarify again: If x represents an individual bit in a long word, the following diagram shows what bits are effected by the different size declarations xxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxx |----------------.l---------------| |-------.w------| |--.b--| There is an instruction that lets you test if a particular bit is set or clear in an internal register or in memory. BTST ---- Syntax btst.s Dn, or btst.s #data, where .s is .b or .l ( note .w not allowed ) If you are testing a bit in memory then the instruction must use the .b size. This instruction test the state of a bit in the destination and sets the zero flag accordingly. You specify the bits reference number either in a data register or as an immediate value. It is usual to follow this instruction with a Bcc instruction so that some action can be taken depending on the state of the bit tested. By way of an example consider the memory address $BFE001. This address is used by the system to indicate various conditions. In particular, bit 6 is used to indicate the state of the left mouse button. If bit 6 is set the the left mouse button is not being pressed, if it is clear then the button is being pressed. By testing this bit a loop can be devised that will wait for the user to press the left mouse button: wait btst #6,$bfe001 bne wait rts The btst instruction is testing bit 6 of the memory address $bfe001. As long as this bit is not zero the program stays in the loop. As soon as the user presses the left mouse button, the program finishes. You could have put the bit number in a data register as follows: moveq.l #6,d0 loop btst d0,$bfe001 bne loop rts If you are wondering about the right mouse button, then test bit 10 of memory address $DFF016 in the same way as for the left mouse button. loop btst #10,$dff016 bne loop rts Boolean Algebra --------------- What ! This is not going to be an indepth study of abstract mathematics, just a brief introduction. Boolean Algebra lends itself very well to computer programming and the 68000 supports all the main Boolean operators AND, OR, NOT and Exclusive OR ( EOR ). Truth Tables ------------ All Boolean operator operate on bits. To see the result of an operator, truth tables are used. AND truth table s d s AND d --------------- 0 0 0 0 1 0 1 0 0 See example 1 1 1 To use this table the combination of bits to be ANDed together should be looked up in the first two coloums. The result of the operation can be found in the third coloum. For example: 1 AND 0 = 0 ( Third row in above example, marked ) Here are the truth tables for the other operators. I will explain how to apply these operations shortly. OR truth table s d s OR d 0 0 0 0 1 1 1 0 1 1 1 1 EOR truth table s d s EOR d 0 0 0 0 1 1 1 0 1 1 1 0 NOT truth table s NOT s 0 1 1 0 Note that NOT operates on only one bit. It changes the state of the bit, so a set bit is cleared and a clear bit is set. The 68000 instructions that correspond to the above operators work on buytes, words or long words. The source and destination must both be the same size and the result is generated in a bitwise fashion. This means that bit 0 of the result is obtained by applying the operator to bit 0 of the source and destination. The same is true for bit 1, bit 2, bit 3 ..... AND --- Syntax AND.s ,Dn or AND.s Dn, where .s can be .b .w or .l Logical AND. This instruction preforms a bitwise AND operation on the source and destination. The result is stored in the destination. Consider ANDing the two words 1011011011110111 and 1100111101011010. This can be done as follows: s = 1 0 1 1 0 1 1 0 1 1 1 1 0 1 1 1 d = 1 1 0 0 1 1 1 1 0 1 0 1 1 0 1 0 s AND d = 1 0 0 0 0 1 1 0 0 1 0 1 0 0 1 0 A program to achieve this would be: move.w #%1011011011110111,d0 move.w #%1100111101011010,d1 and.w d0,d1 display_d1 rts The AND instruction is usually used to mask out any unwanted bits from a value. For instance the DIVU instruction produces a quotient and a remainder in the result. If we are not interested in the remainder we can mask it out by ANDing it with 0's since ANDing with 0 always results in 0. We would need to preserve the quotient however and this could be achieved by ANDing it with 1's since 1 anded with any bit will leave the bit unaffected. The remainder is stored in the high word of the result and the quotient in the low word, so ANDing the result with 000000000000000011111111111111111 will mask out the remainder. Obviously we do not want to be writing long binary values like this in our programs, so we could convert this to hex first. Each set of 4 binary digits can be converted into a hex digit as shown in the table below: binary hex 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F So by writing our binary value into blocks of 4 digits we can easily convert it to hex: 0000 0000 0000 0000 1111 1111 1111 1111 0 0 0 0 F F F F The required hex value will be $0000ffff. Another abreviation we can use is to ommit leading 0's from the value, so $ffff will in fact suffice. Here is an example. To see it in action assemble it to memory using Devpac . Next select the Debug option. You will see three windows. The top window displays the contents of all the internal registers, the bottom left shows the program being debugged and the bottom right shows the contents of memory. If you are not familiar with using this option I suggest you read the manual NOW. Press CTRL z , this executes 1 instruction of your program. Check this by looking at the contents of register d0 in the top window. Now press CTRL z two more times and the division will be preformed. The result can be seen by looking at the contents of register d1 in the top window. See how both quotient ( 2 ) and remainder ( 1 ) are stored in the register. Now press CTRL z again, the remainder has been masked out and is no longer in register d1. Pressing CTRL z once more will return you to Genam. start moveq.l #2,d0 moveq.l #5,d1 divu d0,d1 and.l #$ffff,d1 rts This example can be found on this disc. Do you see how useful the Debug option is for investigating exactly what a program is doing. It is also a priceless tool for finding out where you have made mistakes. I suggest you get you manual out and learn a bit more about Monam, I will answer any questions you may have. ANDI ---- Syntax andi.s #data, Exactly the same as AND except you do not have to use a data register. You can for instance AND an immediate value with data stored in memory. OR -- Syntax or.s ,Dn or or.s Dn, where .s can be .b .w or .l This instruction preforms a bitwise OR of the source on the destination and stores the result in the destination. OR can be used to set bits in the destination. Oring with 0's has no affect on the destination while ORing with 1's will set the corresponding bits in the destination, regardless of their initial state. For example ORing any value with $1 will make the value odd. ORI --- Syntax ori.s #data, Exactly the same as OR except a dtat register does not have to be used. NOT --- Syntax not.s where .s can be .b .w or .l NOT complements all bits in the source and stores the result in the source. Complement means all set bits are cleared and all clear bits are set. eg NOT 00110101 = 11001010 EOR --- Syntax eor.s Dn, where .s can be .b .w or .l This instruction preforms a bitwise Exclusive OR of the source on the destination and stores the result in the destination. This instruction is often used to calculate checksums of data blocks by EORing a starting value with each successive item in the data list. Note that by preforming the same EOR on a value twice, you end up with the value you started with. Assemble and Debug the following example as explained above to see what I mean: start moveq.l #0,d0 move.l d0,d1 move.w #$2a0c,d0 move.w #$5432.d1 eor.w d1,d0 eor.w d1,d0 rts After the second eor instruction, register d0 will again contain the value $2A0C. EORI ---- Syntax eori.s #data, Exactly the same as the EOR instruction except a data register does not have to be used. The End ------- Most of the more commonly used instructions have now been looked at very briefly and it is time to move on. The next article in this series will be concerned with actual programming so you will be able to start writing programs that do something.