Network byte order conversion with “char”

问题

I've always been taught that if an integer is larger than a char, you must solve the byte ordering problem. Usually, I'll just wrap it in the hton[l|s] and convert it back with ntoh[l|s]. But I'm confused why this doesn't apply to single byte characters.

I'm sick of wondering why this is, and would love for a seasoned networks programmer to help me shed some light on why byte orderings only apply for multibyte integers.

Ref: https://beej.us/guide/bgnet/html/multi/htonsman.html

回答1:

What you are looking for is endianness.

A big-endian architecture stores the bytes of a multibyte data type like so:

while a little-endian architecture stores them in reverse:

When data is transferred from one machine to another, the bytes of a single data type must be reordered to correspond with the endianness of the destination machine.

But when a data type only consists of one byte, there is nothing to reorder.

回答2:

Your networking stack will handle the bits inside the bytes correctly, you must only concern yourself with getting the bytes in the right order.

回答3:

Exactly how many ways can you order the bytes in a single char?

回答4:

You need to consider what each function does. From that, you need to apply that knowledge to the size of the type you intend to modify. Consider the following:

#include <stdio.h>
#include <netinet/in.h>

int main () {
    uint16_t i = 42;
    uint8_t c = 42; // a char
    printf ("(uint16_t ) %08X (%d)\n", i, i);
    printf ("(   htons ) %08X (%d)\n", htons(i), htons(i));
    printf ("( uint8_t ) %08X (%c)\n", c, c);
    printf ("(   htons ) %08X (%c)\n", htons(c), htons(c));
    return 0;
}

(uint16_t ) 0000002A (42)
(   htons ) 00002A00 (10752)
( uint8_t ) 0000002A (*)
(   htons ) 00002A00 ()

回答5:

You don't read individual bits off the wire, just bytes. Regardless of the endianness, a single byte is the same backwards and forwards just like the word "I" is the same backwards and forwards.

回答6:

In addition to all the other ways folks have put it: Endianness is about the order of bytes in an integer, not about the order of bits in a byte. The order of bits in a byte is the same even across big-endian and little-endian machines. The only difference is the order in which the bytes themselves are used to store an integer (or short or what-have-you). And so because there's only one byte in a char, there's no difference in how that value is stored, either in a big-endian or little-endian architecture.

回答7:

ezpz code is incorrect..

uint16_t i = 65534;
printf ("(uint16_t ) %08X (%d)\n", i, i);

Returns...
(uint16_t ) 0000FFFE (-2)

You should use a unsigned int instead so it's not interpreted as a signed int

uint16_t i = 65534;
printf ("(uint16_t ) %08X (%u)\n", i, i);

Returns...
(uint16_t ) 0000FFFE (65534)

来源：https://stackoverflow.com/questions/2357720/network-byte-order-conversion-with-char

标签

c++

network-programming