Why is Int32's maximum value 0x7FFFFFFF?

问题

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?

回答1:

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.

回答2:

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

http://en.wikipedia.org/wiki/Two%27s_complement

回答3:

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.

回答4:

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!)

回答5:

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.

回答6:

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 2ⁿ 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

来源：https://stackoverflow.com/questions/13230041/why-is-int32s-maximum-value-0x7fffffff