Precision Software Applications Silver Collection 1

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185 CHAPTER 18 - PORTS In order to communicate with the outside world, the outside world being your printer, your disk drives, your modem etc., the 8086 uses ports. A port is an address distinct from a memory address where the i/o device can be reached.{1} When IBM set up the first PC, they decided what these port addresses would be and what the function of each bit at each address would be. Any VGA, EGA, CGA or monochrome card has to have it's ports at the same addresses as those set by IBM, and these ports have to do the same thing. Normally, an i/o device has more than one port address. The device not only has to transfer the data, it has to tell the computer whether it is ready, find out if the computer wants to send information, confirm that the transfer was successful etc. COM1 is port addresses 3F8 - 3FF. COM2 is 2F8 - 2FF. The CGA adapter is ports 3D0 - 3DF. In the CGA, port 3D9 sets the different colors. The 8086 writes to it, but can't read it. Port 3DA, it's neighbor, tells certain status information and is read only. All control information for the video cards is passed back and forth through the ports. However, the video card comes into memory 50 times a second to see what it should write to the screen. The characters on the screen don't pass through the ports. The mountain comes to Mohammed. The data does pass through a port on the way to your serial printer. The 8086 sends your printer control information through one port address and sends the data through a different port address. The port instructions can be either with AL or AX. AL and AX are the only registers you can use, AL for a byte transfer and AX for a word transfer. There are two forms to the input instruction: in al, port_address or in al, dx port_address is a constant and is limited to 0 - 255. It is a one byte constant. DX is the only register than can be used with the second form. DX can contain any number from 0 - 65535. The ____________________ 1. There is memory address 0000 and there is port address 0000; they are two entirely different things. You can reach memory address 0000 with the DS:SI pair 0000:0000, but DS:SI can't get to port address 0000. The instruction "out 0, al" writes the contents of al to port address 0000, but the OUT instruction can't get anywhere near memory address 0000. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 186 ______________________ following two pieces of code do the same thing: in al, 155 mov dx, 155 in al, dx This code moves one byte of data from port 155 to register AL. The output instruction is similar. We have: out port_address, al or out dx, al where port_address is a constant 0-255 and DX can contain a number 0 - 65535. out 97, ax moves a word from AX to port addresses 97-98. Notice that both the IN and the OUT instructions follow the DESTINATION, SOURCE ordering found in all the other 8086 instructions. The "in ax, port_address" form is not all that useful. If you look at the addresses for the video ports, for COM1 and COM2, you will see that they are all larger than 255. Also, most equipment can be at several different addresses. A modem can be at COM1, COM2, COM3, or COM4. If the port address is hard coded into the instructions, you need 4 subprograms to handle the 4 different addresses. It is easier to find out where the modem is, then use: modem_data_address dw ? mov dx, modem_data_address in al, dx and mov dx, modem_data_address out dx, al If you aren't planning on writing device drivers this is for background information only, since all standard i/o is done through DOS or BIOS interrupts. One last thing is the parity flag. It has been sitting on the screen with all the other flags. What is it for? When you send information over a modem, there is a large possibility of line interference. If it is text, and an occasional screw up is not too bad, using a parity check may be enough.{2} Parity can be even or odd. Even means that an even number of bits are set to 1; odd means that an odd number of bits are set to 1. This is not whether the number is even or odd, it is whether the number of 1 ____________________ 2. It is NEVER enough if you are transferring binary data or executable files. Chapter 18 - Ports 187 __________________ BITS is even or odd. Here's a short list. NUMBER BINARY PARITY 6 0110 even 7 0111 odd 8 1000 odd 9 1001 even 10 1010 even 8 is an even number but has odd parity, 9 is an odd number but has even parity. 6 has two 1 bits (even), 7 has three 1 bits (odd), 8 has one 1 bit (odd), 9 has two 1 bits (even) and 10 has two 1 bits (even). How does this help us? If there is a chance of 1/100 of screwing up a single bit in a byte, there is only a chance of 1/10,000 of screwing up two bits in a byte. What this means is that of every 100 errors, 99 of them will will be detectable because of a change in parity (by changing a bit from 0 to 1 or from 1 to 0) and only 1 of them will go undetected because two changes will keep the same parity. This is a little obscure, so make sure you understand why before continuing. This doesn't help us much yet, because we don't know what the parity was originally. 9 has even parity, 7 has odd parity. What communications programs do is FORCE the parity. They can either force it even or force it odd. The standard ASCII characters end at 127 - that is, 0111 1111 (7F) is the highest legal number you can transmit. The left bit is unused, so we can use it to force the parity. We are going to force it even, but forcing it odd uses the same technique. The steps are: (1) Find the parity of the byte. (2) If it is even, leave it alone, if it is odd, put a 1 in the left hand bit. The parity is now even. If both the sending and receiving program have agreed on even parity, then on the receiving end, the program: (1) Checks for even parity. If the parity is odd, it is an error. (2) Puts a 0 back in the left hand bit. The original number is restored. We are going to make a loop with both parts and use show_regs to watch the parity flag. Here's the program: ; - - - - - - - - - - ENTER DATA BELOW THIS LINE error_banner db "Whoa! We have a screw up.", 13, 10, 0 The PC Assembler Tutor 188 ______________________ ; - - - - - - - - - - ENTER DATA ABOVE THIS LINE ; - - - - - - - - - - ENTER CODE BELOW THIS LINE mov ax_byte, 0A3h ; al binary mov bx_byte, 0A2h ; unsigned half registers mov cx_byte, 0A2h ; unsigned half registers mov dx_byte, 0A3h ; dl binary lea ax, ax_byte call set_reg_style comm_loop: mov ax, 0 call set_count ; reset count to 0 mov bx, 0 ; clear registers mov cx, 0 mov dx, 0 call show_regs ; (1) ; the sending program call get_unsigned_byte and al, 7Fh ; 0111 1111 - make sure al < 128 mov bl, al ; copy to bl call show_regs_and_wait ; (2) and al, al ; check parity call show_regs_and_wait ; (3) jpe do_nothing or al, 80h ; if parity odd, set left bit do_nothing: and al, al ; check parity for show_regs call show_regs_and_wait ; (4) ; the receiving program mov dl, al ; transmit from al to dl mov cl, dl ; copy to cl call show_regs_and_wait ; (5) and dl, dl jpe zero_left_bit ; is parity even? mov ax, error_banner call print_string zero_left_bit: and dl, 7Fh ; 0111 1111 binary, left bit 0 mov cl, dl ; copy to cl call show_regs_and_wait ; (6) jmp comm_loop ; - - - - - - - - - - ENTER CODE ABOVE THIS LINE First, we set the register style so AL and DL are binary, BL and CL are unsigned. We'll use AL and BL for sending, CL and DL for receiving. Next, we reset the counter to 0 so we can see where we Chapter 18 - Ports 189 __________________ are, and zero AX, BX, CX, and DX. We get a byte and make sure that it is 127 or less,{3} then send a copy to BL. BL stays unchanged for the rest of the loop. We test the parity. and if it is odd, change it. Finally, we send it off to DL. On the receiving end, we test the parity. If it is odd, we print an error message. Then we zero the left hand bit. It is faster to zero the left hand bit than to check to see if it needs to be zeroed, so we do it for all data. The result is put in CL for comparison. AL and DL are shown in binary form so you can count the number of 1 bits in the byte. ____________________ 3. The following could happen if we didn't take this step. We get a number > 128 with odd parity. 131 (1000 0011 83h) is an example. The sending program sees that it is odd parity, so it puts a 1 in the left hand bit. But there is already a 1 in the left hand bit, so the parity doesn't change, it is odd. The receiving program gets the byte, tests it, notices that it is odd, and sends back a transmission error. Since the sending program didn't see anything wrong, it just sends the same byte again. It will never get through correctly. In real life, the sending program would probably test the byte to see if it was greater than 127 and print an error if it was. The PC Assembler Tutor 190 ______________________ SUMMARY POSSIBLE I/O INSTRUCTIONS in ax, constant (= port address 0 - 255) in ax, DX in al, constant (= port address 0 - 255) in al, DX out constant, ax (constant = port address 0 - 255) out DX, ax out constant, al (constant = port address 0 - 255) out DX, al In these instructions DX holds a port address and can be from 0 - 65535. PARITY Parity is whether the number of 1 bits in a byte (or word) is even or odd. The 8086 sets the parity flag after most arithmetic and all logical operations. If parity is even, the flag is 1, if parity is odd, the flag is 0.