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

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 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.

[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);