Connection between the BT instruction and the carry flag CF

Question

I know that for example BT BX, 0 transfers the first bit of the BX register to the carry flag CF.

Isn't the carry flag restricted to only have values of 0 and 1, because it's a flag?

Does BT change the CF value from the first bit of the register even if it doesn't hold 0 or 1?

If someone would be able to write here how it works, it would be great!

Answer 1

Yes, the Carry flag can have one of only two possible values: 0 or 1. So does any bit of any register, for instance the least significant bit Nr.0 of BX. Its value is copied by BT BX,0 to CF; BTW this can be used to test if the number in BX is odd or even:

    BT BX,0
    JC Odd
Even:

Answer 2

BT is the instruction to test if a bit is set(1) or not(0). Hence it can only return two values, one or zero, which fit into a (E)FLAG value that can be either TRUE(1) or FALSE(0).

So BT directly copies this bit into the CarryFlag.
The position of the desired bit is given with the immediate value at the end of the instruction - here it is 0, meaning the lowest bit in BX.

(BT does not change the bit tested. For that purpose, you should use BTC, BTR and BTS.)