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Talking to IO boards inside a computer or microcontroller depends upon the principles used by the microprocessor to address physical devices. On Motorola systems all devices connected to the processor local bus are memory mapped. Therefore it is fairly easier to address by means of a pointer the IO device, since it is just like RAM. In the case of Intel devices, the IO is not memory mapped, it is IO mapped. This causes a minor problem is accessing such devices, since the processor has to use IN, OUT commands and the address, therefore a simple pointer will not do. A non-ANSI function exists on PC's called 'portoutb' and 'portinb' these are used to input and output a byte of data (char) to an IO address (int). The syntax depends upon the compiler used by usually consists of outportb(address,data); data = inportb(address); Care should be taken when porting code from an Intel platform to a Motorola platform since these functions are not standard ANSI and will need to converted for the Motorola Platform. Below is shown an example for a 8255 IO board configured at a base address 0x2b4 #include stdio.h #include dos.h void main() { Date: 26th August 1992 Version 1.0 Function : Reads port A and display it on the screen and portb,c Modifications: Non Standard non portable code PC only */ int CardAddress = 0x2b0, BaseAddress, Port_a; int Port_b, Port_c, ControlRegister; unsigned char ControlWord = 0x90, InputValue; BaseAddress = CardAddress + 4; Port_a = BaseAddress; Port_b = BaseAddress + 1; Port_c = BaseAddress + 2; ControlRegister = BaseAddress + 3; ControlWord = 0x90; outportb(ControlRegister, ControlWord); while (1) { InputValue = inportb(Port_a); outportb(Port_b, InputValue); outportb(Port_c, InputValue); printf("Read from A = %x\n", InputValue); } } Despite these additions NON-ANSI functions they are only DOS commands and hence using Windows NT you will need to write a windows driver, for this you will need C++ and a few books on Windows.
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#A Padovan string P(n) for a natural number n is defined as: P(0) = ‘X’ P(1) = ‘Y’ P(2) = ‘Z’ P(n) = P(n-2) + P(n-3), n>2 where + denotes string concatenation. For a string of
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