1004 is the correct answer. One another way to solve the
problem is convert the decimal to binary then to the Hexa
(machine way!)
4100 (decimal) ->1,00 00,00 00,01 00 (binary) -> 1004 (hexa)
I wonder if the nujmber 4100 was picked because it is only 4
more than the "round number" 4096, which is the typical page
size on most 32-bit MMU paging architectures and whose hex
value is 0x1000. Then just add 4 for 0x1004. (Or maybe I'm
over thinking it.)
THE GIVEN NO IS DECIMAL VALUE(4100) SO, FIRST WE NEED TO
CONVERT THAT VALUE IN TO BINARY VALUE.
THIS IS POSSIBLE BY DIVIDING BY 2 UP TO 0 OR 1.
THEN WE NEED TO SEPARATE THOSE VALUES BY 4 APART.
THEN USING 8421 RULE WE HAVE TO GIVE VALUE.
ie,4100 when divide we get 0001 0000 0000 0100.
the answer is exactly 1004.
can you explain in brief what is "r+" mode in a file... i
know that it si used to read and modify rhe existing
content.... but explalanation about the file pointer in "r+"
mode i wann to know???????????