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cendianness

Convert Little Endian to Big Endian

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 stored in little-endian, here are the binary and hex representations of the number:

‭0001 0010 0011 0100 0101 0110 0111 1000‬

‭12345678‬

In big-endian format I believe the bytes should be swapped, like this:

1000 0111 0110 0101 0100 0011 0010 0001

‭87654321

Is this correct?

Also, the code below attempts to do this but fails. Is there anything obviously wrong or can I optimize something? If the code is bad for this conversion can you please explain why and show a better method of performing the same conversion?

uint32_t num = 0x12345678;
uint32_t b0,b1,b2,b3,b4,b5,b6,b7;
uint32_t res = 0;

b0 = (num & 0xf) << 28;
b1 = (num & 0xf0) << 24;
b2 = (num & 0xf00) << 20;
b3 = (num & 0xf000) << 16;
b4 = (num & 0xf0000) << 12;
b5 = (num & 0xf00000) << 8;
b6 = (num & 0xf000000) << 4;
b7 = (num & 0xf0000000) << 4;

res = b0 + b1 + b2 + b3 + b4 + b5 + b6 + b7;

printf("%d\n", res);
Solution

  • OP's sample code is incorrect.

    Endian conversion works at the bit and 8-bit byte level. Most endian issues deal with the byte level. OP's code is doing a endian change at the 4-bit nibble level.
    Recommend instead:

    // Swap endian (big to little) or (little to big)
uint32_t num = 9;
uint32_t b0,b1,b2,b3;
uint32_t res;

b0 = (num & 0x000000ff) << 24u;
b1 = (num & 0x0000ff00) << 8u;
b2 = (num & 0x00ff0000) >> 8u;
b3 = (num & 0xff000000) >> 24u;

res = b0 | b1 | b2 | b3;

printf("%" PRIX32 "\n", res);

    If performance is truly important, the particular processor would need to be known. Otherwise, leave it to the compiler to optimize.

    [Edit] OP added a comment that changes things.
    "32bit numerical value represented by the hexadecimal representation (st uv wx yz) shall be recorded in a four-byte field as (st uv wx yz)."

    It appears in this case, the endian of the 32-bit number is unknown and the result needs to be store in memory in little endian order.

    uint32_t num = 9;
uint8_t b[4];
b[0] = (uint8_t) (num >>  0u);
b[1] = (uint8_t) (num >>  8u);
b[2] = (uint8_t) (num >> 16u);
b[3] = (uint8_t) (num >> 24u);

    [2016 Edit] Simplification

    ... The type of the result is that of the promoted left operand.... Bitwise shift operators C11 §6.5.7 3

    Using a u after the shift constants (right operands) results in the same as without it.

    b3 = (num & 0xff000000) >> 24u;
b[3] = (uint8_t) (num >> 24u);
// same as 
b3 = (num & 0xff000000) >> 24;
b[3] = (uint8_t) (num >> 24);