Question

I ran the following program on little-endian [LE] machine [Linux, Intel processor]. I am unable to explain the 3 outputs in below code snippet. Since machine is LE, the value of a is stored as 0x78563412. When printing, it is displaying its actual value. Since its an LE machine, I expect ntohl() to be a no-op and display 0x78563412, which it is doing. However, I expect 0x12345678 for 2nd print statement containing htonl(). Can someone please help me understand why they are same?

int main() 
{
    int a = 0x12345678; 

    printf("Original - 0x%x\n", (a)); 
    printf("Network - 0x%x\n", htonl(a)); 
    printf("Host - 0x%x\n", ntohl(a)); 

    return 0;
}

Output:

Original - 0x12345678
Network - 0x78563412
Host - 0x78563412

Answer 1

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