问题
I'd like to write a function, in C, which takes the MSB of uint8_t, and if it's set, returns 0xFF and if not 0x00. In short, which returns an integer where all the bits are set to the same value as the MSB.
But I'd like to do it in a completely constant time way, no branches, no array offsets, just mathematical operations which are guaranteed to always touch the same number of bits. And ideally, without any undefined behavior. How can this be done?
回答1:
How about:
#define uint8_msb_to_all_bits(x) (0xFF * ((x) >> 7))
or even better:
#define uint8_msb_to_all_bits(x) (-((x) >> 7))
The way these both work is that, if x is an 8-bit unsigned integer, then x >> 7 is 1 if the MSB of x is set, and 0 otherwise. All that remains is then mapping 1 to 0xFF, which can be done either by multiplication, or, in this particular case, simply by negating the number.
(Yes, negating an unsigned number is well defined in C.)
回答2:
What about
- (x >> 7)
?
Only the MSB is retained and math negation is used to replicate it to all bits.
回答3:
Given the declaration of your uint8_t:
uint8_t x = // MSB is either 1 or 0 if `x` depending on the signed value of `x`
A cheat that assumes 2's complement for signed integers:
return (((uint16_t)(int8_t)x) >> 8) & 0xff;
来源：https://stackoverflow.com/questions/20059480/how-to-write-a-constant-time-function-to-copy-the-most-significant-bit-to-all-bi