Simulating bitwise AND operator

[Puzzle]

Hello,

I have a problem where I must get value of a given binary bit in decimal number:

f(decimal_number, bit_position) = bit_value

For example, getting bit values in decimal 14 (binary 1110):

f(14, 0) = 0
f(14, 1) = 1
f(14, 2) = 1
f(14, 3) = 1

In general it's easy:

(14 & 2⁰) >> 0 = 0
(14 & 2¹) >> 1 = 1
(14 & 2²) >> 2 = 1
(14 & 2³) >> 3 = 1

& is bitwise AND
>> is bitwise arithmetic shift right

But in specific (technical) application I cannot use bitwise operators :(
So far I managed to get rid of one of them:

(14 & 2⁰) / (2⁰) = 0
(14 & 2¹) / (2¹) = 1
(14 & 2²) / (2²) = 1
(14 & 2³) / (2³) = 1

But how to get rid of bitwise AND operator?

Specific application limits me to following things to play with:
+, -, *, /, (, ), square root, x^y, e^x, ln, log, abs, sin, cos, tan, asin, acos, atan
using floating point decimal numbers. And it has to be "single line" - cannot divide solution
to parts, use variables besides given decimal_number or use any if/else etc logic.

It does not have to be "nice" solution, any "hack" will do as long as it works. For example
"helper" constants (specific to bit position) could be used as additional input to the function.

 

[NascentOxygen]

You may be able to adapt the "keep dividing by 2" method for converting decimal to binary.

To see whether there is an 8 (i.e., ––1–––) divide your integer by 16. If the fractional part is >.5 the 3rd bit is a 1, i.e., a 1 in the 2³ column.

 

[rcgldr]

If this is unsigned integer math, using 32 bit numbers, with no overflow check:

d = decimal number
b = bit number

f(d,b) = (d × 2^31-b) / (2³¹)

 

[Puzzle]

f(d,b) = (d × 2^31-b) / (2³¹)

Thats a good way yes, but sadly seems that my "black box" is operating using high precision floating point values :( I think I could still implement what I need if one last obstacle could be overcome:

How to do modulo operation (reminder of division) with toolset that I mentioned?

 

[rcgldr]

f(d,b) = (d × 2^31-b) / (2³¹)

Thats a good way yes, but sadly seems that my "black box" is operating using high precision floating point values.

I missed the part about floating point decimal numbers.

How to do modulo operation (reminder of division) with toolset that I mentioned?

I don't see how this could be done without a floor() function. You mentioned decimal point floating numbers, which could mean BCD (binary coded decimal) strings used for "packed decimal" in Cobol. This would be a problem. If the floating point numbers were binary based, and you knew the format and how hardware handled truncation, you could add, then subtract, a large power of 2 to remove the lower bits, but that still leaves upper bits to be cleared.