Binary Conversion of a File

Question

Hi Guys,
The entire issue is on File Handling.
There is one file say input.dat and another file say output.dat
There are two classes Head and Tail which are Serializable. I am creating objects of those two classes.
I am inserting the first object of Head, then first few bytes from input.dat and then the first object of Tail.
Like this goes on till it reaches to the end of file. Till here everything is fine.
Now the output.dat which is generating, I want to convert it into binary format (e.g. 10110)
How can I go about it? After making it in a binary form, how can I retrive the data ?
Please help me out
Regards
Unmesh

Answer 1

1) the size of the byte array buffer is 4096. Why is it so ?It is an arbitrary value just plucked, almost, out of thin air. It can really be any value that you want, but, since computers a binary machines, a power of two is very desirable (in this case, 2^12). Secondly, in terms of buffer sizes, as the buffer size increases you eventually wind up with a diminishing returns issue (e.g., the larger the buffer, the more memory, more management, more time). To best see this, take a large file and start out with a buffer size of 1 byte and time how long it takes to process the entire file. Do this again but double the buffer size, now 2. It should take roughly half as long to process the entire file. Repeat this process until you see that the processing time comes close to hitting a plateau. Most implementations that utilize buffers tend to use 4K and 8K simply because it just seems to be the best payoff with the smallest price.
2) i couldn't understand this line of the code
for (int mask = 128; mask > 0; mask /= 2)
From where are you getting these numbers 128 and 2 ? why are you dividing mask by 2 ?A byte consists of 8 bits (to be more precise, let us call this an octet). Each bit has a value - basic number system thing. So, for an octet, the bit values are (most significant bit first):
7   | 6  | 5  | 4  | 3  | 2  | 1  | 0          bit
128 | 64 | 32 | 16 | 8  | 4  | 2  | 1          decimal bit value
80  | 40 | 20 | 10 | 08 | 04 | 02 | 01         hex bit valueTo liken this to the decimal, base 10, system, bit 0 is the 1's column, bit 1 is the 10's column, bit 2 is the 100's column and so on. The binary system (base 2) is mathematically the same. It is generalized like this for a given value at a given "column" for any number system:
value := b^i
where:
b is the base number system in question
i is the column in question
So, to find the complete value of any number in any number system:
value := db^i + db^i-1 + db^i-2 + ... + db^1 + db^0
where:
b is the base number system in question
i is the column in question
d is the value of the digit in the column in question
I should also note that jummoMa1's approach of using bit shifting is more efficient.