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@int_main_void

ビットマスク

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Python - 当初版

bitmask.py
# b5-b3のマスクを得る. 
startbit=3   # startbitをb3
bitlen=3     # startbitを含めて3bitのビット長(b5b4b3)
b=0
for i in range(startbit, startbit + bitlen):
     b|=1<<i

bin(b) #=> '0b111000'

# b10-b0のマスクを得る. 
startbit=0 # startbitをb0
bitlen=11 # startbitを含め11bitのビット長(b10...b1b0)
b=0
for i in range(startbit, startbit + bitlen):
    b|=1<<i

bin(b) # '0b11111111111'

# b49-45のマスクを得る.
startbit=45 # startbitをb45
bitlen=5 # startbitを含め5bitのビット長(b49b48b47b46b45)
b=0
for i in range(startbit, startbit + bitlen):
    b|=1<<i

bin(b) #=> '0b11111000000000000000000000000000000000000000000000'

Python - shiracamusさん版

bitmask.py
def bitmask(startbit, bitlen):
     return ((1 << bitlen) - 1) << startbit

startbit=3, bitlen=3で具体化すると、補数的な感じでマスクが求められていることがわかる
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

C - 当初版

  • bitmask(0,64)(b63-b0のbitmask)が作成できないバグあり(コメント参照)

bitmask.c
#include <stdio.h>
unsigned long long bitmask1(
    unsigned char startbit,
    unsigned char bitlen
){
    return (((unsigned long long)0x1 << bitlen)-0x1) << startbit;
}

C - shiracamusさん版

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;
}

C - 第3版

いろいろ考えた結果、あまり面白くない感じになりました。
* 間違いがあったため、修正(bm %= 64u; → bitlen %= 64u;)

実行結果比較
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 ;
}
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