A good answer might be:

Two bytes. For ease in doing IO, booleans are represented using one byte (look at the documentation).

The documentation for several of the methods uses the terms "low byte" and "high byte". These are used when a data type uses more than one byte. The low byte is the byte that holds the least significant part of an integer. If you think in terms of writing a bit pattern on paper, the low byte is the rightmost eight bits. A short holds a 16-bit pattern such as:

01001010 00001111

The low order byte is 00001111. (The space between the groups is there for readability). For integer types, the low order byte holds the part of the number that consists of powers of two from 0 to 7. An integer value from 0 to 255 will fit into just the low byte. For example, here is the bit pattern for an int that holds the value 98:

00000000 00000000 00000000 01100010

If this int is written out with writeInt(), the bytes are written from high byte to low byte (or from left to right in the above pattern).

For the most part you don't need to think about this unless you are doing serious systems programming. A course in assembly language or computer architecture will say much more about bit patterns.

QUESTION 11:

How many bytes are written by the following:

out.writeInt( 0 );
out.writeDouble( 12.45 );