Closed JachinJiang closed 10 months ago
tmp = (tmp2 >> 30) | ((tmp1 << 1) & 0x6);
to construct a 3-bit number, tmp1 contributes 2bit, why tmp2 also contributes 2bit? I'm confused.
Hi, tmp2 was already shifted by one before in this line and thus only contributes 1 bit in this expression.
tmp2
oh, my fault, I'll close this issue, thank you.
tmp = (tmp2 >> 30) | ((tmp1 << 1) & 0x6);
to construct a 3-bit number, tmp1 contributes 2bit, why tmp2 also contributes 2bit? I'm confused.