Generating a byte with an array of bit positions?-CodePudding

Let's say I have a list of numbers:

[1, 2, 5]

I need to generate a byte that represents the 5, 2 and 1 "position on". Which will result in the byte of value 19.

00010011

Is there an easy way to do this that doesn't involve me playing with an array of ones and zeros and them trying to convert it to a byte?

CodePudding user response：

Doing 2**(n-1) will set a bit in the position you want:

print(bin(2**(1-1))) # 00001
print(bin(2**(2-1))) # 00010
print(bin(2**(5-1))) # 10000

This is equivalent to bit shifting:

1<<(1-1) # 00001
1<<(2-1) # 00001
1<<(5-1) # 10000

You can sum 2**(n-1) for each number in your list:

nums = [1, 2, 5]

print(sum(2**(x-1) for x in nums))

Which results in the same result in applying the 'and' operation on each of the binary numbers:

00001
00010 AND
10000 AND
-----
10011

Which is equal to 19.

You can convert this to binary using the bin function:

print(bin(sum(2**(x-1) for x in nums)))

Output:
0b10011