Extending Fast numeric string to int to converting hexadecimal number strings is easy.

The main wrinkle is that the range of digits expands from the original '0'-'9' (0x30-0x39) to '0'-'9', 'A'-'F', 'a'-'f' (0x30-0x39, 0x41-0x46, 0x61-0x66)

The high alphabetic hexadecimal digits ('A'-'F', 'a'-'f' 0x41-0x46, 0x61-0x66) are not contiguous with the lower decimal digits ('0'-'9' 0x30-0x39).

If the code masks off the high nibble (4 bits) as it did when processing numeric strings, the code would be left with hexadecimal digits in the range 1-6, 0x01-0x06, when it should be in the range 10-15, 0x0A-0x0F.

So the code needs to add 9 to ONLY the hexadecimal digits in the range 0x41-0x46 and 0x61-0x66.

Concept #1

Notice that all of the high alphabetic hexadecimal digits have the 0x40 bit set.

The code can key off of the 0x40 bit set, to locate the alphabetic hexadecimal digits.

Shift that bit down to the 1 bit and multiply by 9 to create the number 9 for the appropriate byte.

Then we add to the hex digit, toggle the 0x40 bit and we should be off to the races with hex digits in the range 0x0-0xF.

Code needs slight tweaks to the multiplication constants, since we’re using base 16 now, not base 10 for the digits.

Here’s the final algorithm:

typedef unsigned long long ULL;
ULL n = (*(ULL *)(buffer)) & 0x4F4F4F4F4F4F4F4Full;

ULL alphahex = (ULL)(n & 0x4040404040404040ull);
// ULL  nine = (alphahex >> 6) * 9;
// ULL n0 = alphahex == 0 ? n : nine + (n ^ alphahex);
ULL n0 = alphahex == 0 ? n 
                       : ((alphahex >> 6) * 9) + (n ^ alphahex);
// 0x1001 == 4097 == 256 * 16 + 1
ULL n1 = n0 * 0x1001 >> 8;
// 0x1000001 == 16777217 == 65536 * 256 + 1
ULL n2 = (n1 & 0x00FF00FF00FF00FFull) * 0x1000001 >> 16;
// 0x1000000000001 == 281474976710657 == 4294967296 * 65536 + 1
unsigned long num = (n2 & 0x0000FFFF0000FFFFull) * 0x1000000000001 >> 32;