Answer Posted / uttam kumhar

1 char always hold 1 byte in c in any 32-bit or 64-bit system
but in case of int it vary on the system.if there is 32-bit system the int hold 4-byte memory ...and if 64-bit system int hold 2-byte memory....

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how can i access a direct (absolute, not the offset) memory address? here is what i tried: wrote a program that ask's for an address from the user, creates a FAR pointer to that adress and shows it. then the user can increment/decrement the value in that address by pressing p(inc+) and m(dec-). NOW, i compiled that program and opened it twice (in 2 different windows) and gave twice the same address to it. now look what happen - if i change the value in one "window" of the program, it DOES NOT change in the other! even if they point to the same address in the memory! here is the code snippet: //------------------------------------------------------ #include //INCLUDE EVERY KNOWN HEADER FILE #include //FOR ANY CASE... #include #include #include main() { int far *ptr; //FAR POINTER!!! long address; char key=0; //A KEY FROM THE KEYBOARD int temp=0; clrscr(); cout<<"Enter Address:"; cin>>hex>>address; //GETS THE ADDRESS clrscr(); (long)ptr=address; temp=*ptr; //PUTS THE ADDRESS IN THE PTR cout<<"["<

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