从位置pos开始生成len位设置为1的掩码的最快方法是什么(就常见现代架构的cpu周期而言):
What is the fastest way (in terms of cpu cycles on common modern architecture), to produce a mask with len bits set to 1 starting at position pos:
template <class UIntType>
constexpr T make_mask(std::size_t pos, std::size_t len)
{
// Body of the function
}
// Call of the function
auto mask = make_mask<uint32_t>(4, 10);
// mask = 00000000 00000000 00111111 11110000
// (in binary with MSB on the left and LSB on the right)
另外，是否有任何编译器内在函数或 BMI 函数可以提供帮助?
Plus, is there any compiler intrinsics or BMI function that can help?
如果用starting at pos"表示掩码的最低位在2对应的位置pos(如你的例子):
If by "starting at pos", you mean that the lowest-order bit of the mask is at the position corresponding with 2pos (as in your example):
((UIntType(1) << len) - UIntType(1)) << pos
如果 len 可能是 ≥UIntType 中的位数，通过测试避免未定义行为:
If it is possible that len is ≥ the number of bits in UIntType, avoid Undefined Behaviour with a test:
(((len < std::numeric_limits<UIntType>::digits)
? UIntType(1)<<len
: 0) - UIntType(1)) << pos
(如果 pos 也有可能是 ≥ std::numeric_limits
(If it is also possible that pos is ≥ std::numeric_limits<UIntType>::digits, you'll need another ternary op test.)
您也可以使用:
(UIntType(1)<<(len>>1)<<((len+1)>>1) - UIntType(1)) << pos
以三个额外的移位运算符为代价避免了三元运算；我怀疑它是否会更快，但需要仔细的基准测试才能确定.
which avoids the ternary op at the cost of three extra shift operators; I doubt whether it would be faster but careful benchmarking would be necessary to know for sure.
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