What is the implementation of GCC's (4.6+) __builtin_clz
? Does it correspond to some CPU instruction on Intel x86_64 (AVX)
?
Yes, and no.
CLZ (count leading zero) and BSR (bit-scan reverse) are related but different. CLZ equals (type bit width less one) - BSR. CTZ (count trailing zero), also know as FFS (find first set) is the same as BSF (bit-scan forward.)
Note that all of these are undefined when operating on zero!
In answer to your question, most of the time on x86 and x86_64, __builtin_clz generates BSR operation subtracted from 31 (or whatever your type width is), and __builting_ctz generates a BSF operation.
If you want to know what assembler GCC is generating, the best way to know is to see. The -S flag will have gcc output the assembler it generated for the given input:
gcc -S -o test.S test.c
Consider:
unsigned int clz(unsigned int num) {
return __builtin_clz(num);
}
unsigned int ctz(unsigned int num) {
return __builtin_ctz(num);
}
On x86 for clz gcc (-O2) generates:
bsrl %edi, %eax
xorl $31, %eax
ret
and for ctz:
bsfl %edi, %eax
ret
Note that if you really want bsr, and not clz, you need to do 31 - clz (for 32-bit integers.) This explains the XOR 31, as x XOR 31 == 31 - x (this identity is only true for numbers of the from 2^y - 1) So:
num = __builtin_clz(num) ^ 31;
yields
bsrl %edi, %eax
ret