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How many bits would you need to store a positive integer for example in the billions? Would you have to use the log2 N to find this out?
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1 Integers. Integers are commonly stored using a word of memory, which is 4 bytes or 32 bits, so integers from 0 up to 4,294,967,295 (232 - 1) can be stored.
With 4 bits, the maximum possible number is binary 1111 or decimal 15. The maximum decimal number that can be represented with 1 byte is 255 or 11111111. An 8-bit word greatly restricts the range of numbers that can be accommodated.
A single bit can only represent two different values.
With 4 bits, it is possible to create 16 different values. All single-digit hexadecimal numbers can be written with four bits. Binary-coded decimal is a digital encoding method for numbers using decimal notation, with each decimal digit represented by four bits.
Yes. the maximal number stored in k bits is 2^k-1, since there is 2^k options for the bits, and one of them is zero.
Therefore, the number of bits required to store a number N is log2(N), but since there is no half bit, you need to round it up to the cloest integer above.
Note: if you need to include negative numbers, there must be one more bit for the sign.
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Since I've seen the answer reported incorrectly so many times, I thought I would post the correct answer.
The number of bits needed to represent the positive integer n is
bits = floor( log2(n) + 1 )
where log2 means log base 2.
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