why unsigned int 0xFFFFFFFF is equal to int -1?

原帖: http://stackoverflow.com/questions/1863153/why-unsigned-int-0xffffffff-is-equal-to-int-1

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perhaps it's a very stupid question but I'm having a hard time figuring this out =)

in C or C++ it is said that the maximum number a size_t (an unsigned int data type) can hold is the same as casting -1 to that data type. for example see http://stackoverflow.com/questions/1420982/invalid-value-for-sizet

Why?? I'm confused..

I mean, (talking about 32 bit ints) AFAIK the most significant bit holds the sign in a signed data type (that is, bit 0x80000000 to form a negative number). then, 1 is 0x00000001.. 0x7FFFFFFFF is the greatest positive number a int data type can hold.

then, AFAIK the binary representation of -1 int should be 0x80000001 (perhaps I'm wrong). why/how this binary value is converted to anything completely different (0xFFFFFFFF) when casting ints to unsigned?? or.. how is it possible to form a binary -1 out of 0xFFFFFFFF?

I have no doubt that in C: ((unsigned int)-1) == 0xFFFFFFFF or ((int)0xFFFFFFFF) == -1 is equally true than 1 + 1 == 2, I'm just wondering why.

thanks!

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C and C++ can run on many different architectures, and machine types. Consequently, they can have different representations of numbers: Two's complement, and Ones' complement being the most common. In general you should not rely on a particular representation in your program.

For unsigned integer types (size_t being one of those), the C standard (and the C++ standard too, I think) specifies precise overflow rules. In short, if SIZE_MAX is the maximum value of the type size_t, then the expression

(size_t) (SIZE_MAX + 1)

is guaranteed to be 0, and therefore, you can be sure that (size_t) -1 is equal to SIZE_MAX. The same holds true for other unsigned types.

Note that the above holds true:

• for all unsigned types,
• even if the underlying machine doesn't represent numbers in Two's complement. In this case, the compiler has to make sure the identity holds true.

Also, the above means that you can't rely on specific representations for signed types.

Edit: In order to answer some of the comments:

Let's say we have a code snippet like:

int i = -1;long j = i;

There is a type conversion in the assignment to j. Assuming that int and long have different sizes (most [all?] 32-bit systems), the bit-patterns at memory locations for i and j are going to be different, because they have different sizes. The compiler makes sure that the values of i and j are -1.

Similarly, when we do:

size_t s = (size_t) -1

There is a type conversion going on. The -1 is of type int. It has a bit-pattern, but that is irrelevant for this example because when the conversion to size_t takes place due to the cast, the compiler will translate thevalue according to the rules for the type (size_t in this case). Thus, even if int and size_t have different sizes, the standard guarantees that the value stored in s above will be the maximum value that size_t can take.

If we do:

long j = LONG_MAX;int i = j;

If LONG_MAX is greater than INT_MAX, then the value in i is implementation-defined (C89, section 3.2.1.2).

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It's called two's complement. To make a negative number, invert all the bits then add 1. So to convert 1 to -1, invert it to 0xFFFFFFFE, then add 1 to make 0xFFFFFFFF.

As to why it's done this way, Wikipedia says:

The two's-complement system has the advantage of not requiring that the addition and subtraction circuitry examine the signs of the operands to determine whether to add or subtract. This property makes the system both simpler to implement and capable of easily handling higher precision arithmetic.

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这个一样可以参考, 深入理解计算机系统第 2 章, 讲解的很详细, 下溢和上溢等.