Why does the range of int has a minus 1?

Question

I read that the range of an int is dependent on a byte.

So taking int to be 4 bytes long, thats 4 * 8 bits = 32 bits.

So the range should be : 2 ^ (32-1) = 2 ^ (31)

Why do some people say its 2^31 - 1 though?

Thanks!

Answer 1

Because the counting starts from 0

And the range of int is 2,147,483,647 and 2^32 which is 2,147,483,648. hence we subtract 1

Also the loss of 1 bit is for the positive and negative sign

Check this interestinf wiki article on Integers:-

The most common representation of a positive integer is a string of bits, using the binary numeral system. The order of the memory bytes storing the bits varies; see endianness. The width or precision of an integral type is the number of bits in its representation. An integral type with n bits can encode 2n numbers; for example an unsigned type typically represents the non-negative values 0 through 2n−1. Other encodings of integer values to bit patterns are sometimes used, for example Binary-coded decimal or Gray code, or as printed character codes such as ASCII.

There are four well-known ways to represent signed numbers in a binary computing system. The most common is two's complement, which allows a signed integral type with n bits to represent numbers from −2(n−1) through 2(n−1)−1. Two's complement arithmetic is convenient because there is a perfect one-to-one correspondence between representations and values (in particular, no separate +0 and −0), and because addition, subtraction and multiplication do not need to distinguish between signed and unsigned types. Other possibilities include offset binary, sign-magnitude, and ones' complement.

Answer 2

You mean 2³²-1, NOT 2^32-1.

But your question is about why people use 2³¹. The loss of a whole bit is if the int is a signed one. You lose the first bit to indicate if the number is positive or negative.

A signed int (32 bit) ranges from -2,147,483,648 to +2,147,483,647. An unsigned int (32 bit) ranges from 0 to 4,294,967,295 (which is 2³² -1).