Why do not need to worry about Big and little Endian in Java

Question

Please help, I do not understand this concept please explain.
The architecture of a CPU is either little-endian or big-endian; some modern CPU's allow a choice via software.
The TCP/IP protocol standard specifies that all the bytes that make up an item must be sent in "network order", which happends to be big-endian. Intel Pentium CPU's are little-endian.
This implies that on an Intel machine the TCP software will have to chop an int into bytes and then reverse the bytes before transmitting them.
Why does the JAVA TCP software does not need to perform the reversal?
Thanks,
Alex

Answer 1

Java doesn't give you direct access to the individual bytes of a larger data item such as an integer. For this reason you don't have the usual endian problems that occur in C. The actual handling of this is in the DataOutputStream and DataInputStream where the integer is coverted to and from bytes using arithmetic, not by fiddling with the internal structure.
Note that regardless of the machine architecture the operation value%256 will return the low order 8 bits. It's less efficient than assigning an int* to a char*, but it's not fraught with the endian problems or any of the other hardware baggage.