Convert char from big endian to little endian in C

问题:

I'm trying to convert a char variable from big endian to little endian.

Here it is exactly:

char name[12];

I know how to convert an int between big and little endian, but the char is messing me up.

I know I have to convert it to integer form first, which I have.

For converting an int this is what I used:

(item.age >> 24) | ((item.age >> 8) & 0x0000ff00) | ((item.age << 8) & 0x00ff0000) | (item.age << 24);

For converting the char, I'd like to do it in the same way, if possible, just because this is the only way I understand how to.

回答1:

Endianess only affects byte order. Since char is exactly one byte it is the same on all endianess formats.

And as a note: you may want to look up the endianess conversion functions in libc:

htonl, htons
ntohl, ntohs

回答2:

Byte-order (a.k.a. Endian-ness) has no meaning when it comes to single-byte types.

Although char name[12] consists of 12 bytes, the size of its type is a single byte.

So you need not do anything with it...

BTW, the size of int is not necessarily 4 bytes on all compilers, so you might want to use a more generic conversion method (for any type):

char* p = (char*)&item;
for (int i=0; i<sizeof(item)/2; i++)
{
    char temp = p[i];
    p[i] = p[sizeof(item)-1-i];
    p[sizeof(item)-1-i] = temp;
}