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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.
How many address bits are required to represent a 32 K memory ? Ans. 32K = 25 x 210 = 215, Hence 15 address bits are needed; Only 16 bits can address this.
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