On 2020/2/28 11:33, Richard Henderson wrote:
On 2/27/20 5:50 PM, LIU Zhiwei wrote:
This is not what I had in mind, and looks wrong as well.
int idx = (index * mlen) / 64;
int pos = (index * mlen) % 64;
return (((uint64_t *)v0)[idx] >> pos) & 1;
You also might consider passing log2(mlen), so the multiplication could be
strength-reduced to a shift.
I don't think so. For example, when mlen is 8 bits and index is 0, it will
reduce to
return (((uint64_t *)v0)[0]) & 1
And it's not right.
The right bit is first bit in vector register 0. And in host big endianess,
it will be the first bit of the seventh byte.
You've forgotten that we've just done an 8-byte big-endian load, which means
that we *are* looking at the first bit of the byte at offset 7.
It is right.
Yes, that's it.
You don't need to pass mlen, since it's
Yes.
I finally remembered all of the bits that go into mlen and thought I had
deleted that sentence -- apparently I only removed half. ;-)
r~
