bitmask.py
# b5-Get the mask of b3.
startbit=3 #startbit b3
bitlen=3 #Bit length of 3 bits including start bit(b5b4b3)
b=0
for i in range(startbit, startbit + bitlen):
b|=1<<i
bin(b) #=> '0b111000'
# b10-Get the mask of b0.
startbit=0 #startbit b0
bitlen=11 #Bit length of 11 bits including start bit(b10...b1b0)
b=0
for i in range(startbit, startbit + bitlen):
b|=1<<i
bin(b) # '0b11111111111'
# b49-Get 45 masks.
startbit=45 #startbit b45
bitlen=5 #Bit length of 5 bits including start bit(b49b48b47b46b45)
b=0
for i in range(startbit, startbit + bitlen):
b|=1<<i
bin(b) #=> '0b11111000000000000000000000000000000000000000000000'
bitmask.py
def bitmask(startbit, bitlen):
return ((1 << bitlen) - 1) << startbit
If you embody it with startbit = 3 and bitlen = 3, you can see that the mask is required with a complement feeling. 1 << bitlen : 1 <<3 = 0b1000
(1 << bitlen) - 1 : (1 << 3) - 1 = 0b1000 - 1 = 0b111
(1 << bitlen) - 1) << starbit : ((1 << 3) - 1) << 3 = 0b111 << 3 = 0b111000
http://melpon.org/wandbox/permlink/KAdTUEdphWrNncP0
bitmask.c
#include <stdio.h>
unsigned long long bitmask1(
unsigned char startbit,
unsigned char bitlen
){
return (((unsigned long long)0x1 << bitlen)-0x1) << startbit;
}
bitmask2.c
unsigned long long bitmask2(
unsigned char startbit,
unsigned char bitlen
){
if (startbit >= 64) return 0;
if (bitlen >= 64) return ~0ULL << startbit;
return ((1ULL << bitlen) - 1ULL) << startbit;
}
As a result of various thoughts, it became less interesting.
Execution result comparison http://melpon.org/wandbox/permlink/hTrTCCmfgjORuGsT
bitmask3.c
unsigned long long bitmask3(
unsigned char startbit,
unsigned char bitlen
){
unsigned long long bm = 0x0UL;
if (bitlen == 64u){
bm = ~(0x0ULL);
} else {
bitlen %= 64u;
bm = (0x1ULL << bitlen) - 0x1ULL ;
}
startbit %= 64u;
bm <<= startbit;
return bm ;
}
Recommended Posts