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);
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);