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 ;
}
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