Confusion in htons- little endian/ big endian

Question

When I send a integer variable from one process to other through socket, and then printing the value at received end, the value is still the same without using ntohl/htonl, then where do I need to use these functions other than initializing socket structures. I understand litte/big endian. But why do we need to convert port and IP nos to host/network byte order when value remains the same. Please explain in detail how the integer is tranferred over network?

Answer 1

If you want your program to be portable, then any time you send an integer greater than 1 byte in size over the network, you must first convert it to network byte order using htons or htonl, and the receiving computer must convert it to host byte order using ntohs or ntohl.

In your case, the reason the value is still the same is probably because the sending computer and the receiving computer are of the same endianness. In other words, the sending computer and the receiving computer you're working with are both little endian (or big endian, whatever the case may be.)

But if you want your program to be portable, you can't rely on this to always be the case. One day, for example, the sending computer may be an Intel x86, and the receiving may be a Sun SPARC, and then your program will fail if you don't use htons.