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Write a program to input an address i.e. 20 Hex and read the byte (char) contents of that address
Answer include void main() { char prompt; Date: 26th August 2012
Version 1.0 Function : To read an address in memory Modifications: none*/ unsigned long int address; unsigned char *add; unsigned char data; printf("Please enter in the address you want to read\n\r"); scanf("%lx",&address); add = (unsigned char *)address; data = *add; printf("The data at the address %08lx is %02x\n\r",address,data); printf("Press and key to exit \n\r"); scanf("\n%c",&prompt); } Note: Care should be taken since some operating systems do not allow the user to read any memory address. The PC environment is one of these in that the PC runs under protected memory and hence looking at memory outside this area can yield 'General protection errors '. In order to access the memory in a PC, the operating system must tag the memory area for common use by the program. This is usually achieved by either writing a windows driver for the hardware/address or my poking in an allocated area i.e. allocating static space by an array or using malloc to reserve bytes of area.
The commands (instructions) are native instructions to the 68HC11 and therefore it is termed a low level language program. Examining the program the instructions (mnemonics) they a
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Address translation: Compiler time : If it is known in advance that a program will reside at a particular location of primary memory, and then the compiler can be told to
The 68HC11 series is based on the Motorola 6800/1 programming instruction set and hence is a fairly simple 8 bit microprocessor. The internal structure of the 6800/1 is shown below
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END DIRECTIVE: ENDS directive ends a segment and ENDP directive ends a procedure and END directive ends whole program which appears as last statement. ASSUME Directive: An .
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