Convert Little Endian to Big Endian

Question

I just want to ask if my method is correct to convert from little endian to big endian, just to make sure if I understand the difference.

I have a number which is st

Answer 1

You can use the lib functions. They boil down to assembly, but if you are open to alternate implementations in C, here they are (assuming int is 32-bits) :

void byte_swap16(unsigned short int *pVal16) {

//#define method_one 1
// #define method_two 1
#define method_three 1
#ifdef method_one
    unsigned char *pByte;

    pByte = (unsigned char *) pVal16;
    *pVal16 = (pByte[0] << 8) | pByte[1];
#endif

#ifdef method_two
    unsigned char *pByte0;
    unsigned char *pByte1;

    pByte0 = (unsigned char *) pVal16;
    pByte1 = pByte0 + 1;
    *pByte0 = *pByte0 ^ *pByte1;
    *pByte1 = *pByte0 ^ *pByte1;
    *pByte0 = *pByte0 ^ *pByte1;
#endif

#ifdef method_three
    unsigned char *pByte;

    pByte = (unsigned char *) pVal16;
    pByte[0] = pByte[0] ^ pByte[1];
    pByte[1] = pByte[0] ^ pByte[1];
    pByte[0] = pByte[0] ^ pByte[1];
#endif


}



void byte_swap32(unsigned int *pVal32) {

#ifdef method_one
    unsigned char *pByte;

    // 0x1234 5678 --> 0x7856 3412  
    pByte = (unsigned char *) pVal32;
    *pVal32 = ( pByte[0] << 24 ) | (pByte[1] << 16) | (pByte[2] << 8) | ( pByte[3] );
#endif

#if defined(method_two) || defined (method_three)
    unsigned char *pByte;

    pByte = (unsigned char *) pVal32;
    // move lsb to msb
    pByte[0] = pByte[0] ^ pByte[3];
    pByte[3] = pByte[0] ^ pByte[3];
    pByte[0] = pByte[0] ^ pByte[3];
    // move lsb to msb
    pByte[1] = pByte[1] ^ pByte[2];
    pByte[2] = pByte[1] ^ pByte[2];
    pByte[1] = pByte[1] ^ pByte[2];
#endif
}

And the usage is performed like so:

unsigned short int u16Val = 0x1234;
byte_swap16(&u16Val);
unsigned int u32Val = 0x12345678;
byte_swap32(&u32Val);