What is the upper bound of BigInteger with character array implementation?

Question

If I impement BigInteger with a character array (in C++), in terms of power of 10, what is my upper bound in a 32bit system?

In other words,

- 10^x < N <= 10^x

(first character is reserved for sign).

What is x in 32 bit system?

Please ignore for now that we have reserved memory for OS and consider all 4GB memory is addressable by us.

Answer 1

An 8-bit byte can hold 2 ⁸ , or 256 unique values.

4GB of memory is 2 ³² , or 4294967296 bytes .

Or 4294967295, if we subtract the one byte that you want to reserve for a sign

That's 34359738360 bits .

This many bits can hold 2 ^34359738360 unique values.

- 10^x < N <= 10^x

(first character is reserved for sign).

What is x in 32 bit system?

Wolfram Alpha suggests - 10^1292913986 < N <= 10^1292913986 as the largest representable powers of 10.

So x is 1,292,913,986.

Answer 2

(−(2^(n−1))) to (2^(n−1) − 1) calculates the range of a signed integer where n is the number of bits. [1]

Assuming your referring to the whole 4GB of memory being allocated, that is 2 ³² (4,294,967,295) addressable bytes in 32 bit memory space, which is 2 ³⁵ (34,359,738,368) bits.

Put that into the formula at the start and you get a range of - (2 ^{2 35 -1} ) to 2 ^{2 35 -1} -1

This is assuming you use a bit for a sign, instead of a whole byte. If your going a use a whole byte for sign, you should calculate the unsigned range of 2 ³⁵ -8 bits. Which is from 0 to 2 ^{2 35 -8} −1

According to this page, to convert from an exponent of base 2 to an exponent of base 10, you should use the formula x = m*ln(2)/ln(10),where you are converting from 2 ^m to 10 ^x .

Therefore, your answer is that the upper bound is 10 ^{2 35 -8 *ln(2)/ln(10)} . I'm not going to even attempt to change that exponent into a decimal value.