Far pointer questions with solution

(1) What will be output ? void main() { char far *p=(char far *)0x55550005; char far *q=(char far *)0x53332225; clrscr(); *p=25; (*p)++; printf("%d",*q); getch(); } Output: 26 Explanation: Far address of p and q are representing same physical address . Physical address of 0x55550005= 0x5555*ox10+ox0005= 0x55555 Physical address of 0x53332225=0x5333*0x10+ox2225=0x55555 *p =25, means content at memory location 0x55555 is assigning value 25 (*p)++ means increase the content by one at memory location 0x5555 so now content at memory location 0x55555 is 26 *q also means content at memory location 0x55555 which is 26 (10) what will be output ? #include void main() { unsigned int i,offset_addr,seg_addr,; char far *ptr=(char far *) 5555 FF99; for (i=0;i<300;i++) { seg_addr=FP_SEG(p); offset_addr=FP_OFF(p); printf(“\n seg_addr %x offset_addr %x ”,seg_addr,offset_addr); p++; delay(100); } }