I saw in MSDN documents that the maximum value of `Int32`

is `2,147,483,647`

, hexadecimal `0x7FFFFFFF`

.

I think, if it's `Int32`

it should store 32-bit integer values that finally should be `4,294,967,295`

and hexadecimal `0xFFFFFFFF`

.

My question is why `Int32`

stores 31-bit integer values?

It's because it's a signed integer. An unsigned 32-bit integer give you the value you expect.

Check out this MSDN page - http://msdn.microsoft.com/en-us/library/exx3b86w(v=vs.80).aspx

For a more in depth explanation on why this is check out the link in Jackson Popes answer related to Two's Complement number representation.

Also some further reading.

Because one bit is used to store the sign (Int32 can be less than zero).

Int32 and Int64 are both signed so they can handle integer values from -capacity/2 to (capacity/2)-1 (for zero) that is why the max value isn't the one you expected. But you can get what you want by using an unsigned int to have only positive numbers.

You are not considering the negative numbers.
`Int32`

have the sign.

From MSDN: http://msdn.microsoft.com/en-us/library/system.int32.minvalue.aspx
The `MinValue`

is `-2,147,483,648`

; that is, hexadecimal `0x80000000`

.

The first bit is the sign - an int32 is signed, i.e. it can be positive/negative (well I probably shouldn't say 'first' bit!)

answered on Stack Overflow Nov 5, 2012 by Charleh

In a 2's complement signed n-bit type, the range is from -2^{n-1} to 2^{n-1}-1 because with n bits you can represent 2^{n} different values, half of which is used for signed numbers because of the sign bit. The remaining 2^{n-1} half is used for non-negative number. Since one is used for 0, there are only 2^{n-1}-1 remaining values for positive numbers

answered on Stack Overflow May 30, 2014 by phuclv

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