Same output for htonl() and ntohl() on an integer

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Since its an LE machine, I expect ntohl() to be a no-op

That’s the mistake. Network byte order is big-endian, host byte order is little-endian. Therefore, both ntohl and htonl return a byte-swapped version of their input.

Remember, the point of htonl is that you can take an integer on the host, then write:

int i = htonl(a);

and the result is that the memory of i, when interpreted using network byte order, has the same value that a does. Hence, if you write the object representation of i to a socket and the reader at the other end expects a 4-byte integer in network byte order, it will read the value of a.

and display 0x78563412

Is this what you intended to write? If ntohl were a no-op (or rather, an identity function), then your third line necessarily would print the same thing as your first line, because you would have ntohl(a) == a. This is what happens on big-endian implementations, where your program prints:

Original - 0x12345678
Network - 0x12345678
Host - 0x12345678

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