Why is −1 > sizeof(int)?

问题:

Consider the following code:

template<bool> class StaticAssert;
template<> class StaticAssert<true> {};
StaticAssert< (-1 < sizeof(int)) > xyz1; // Compile error
StaticAssert< (-1 > sizeof(int)) > xyz2; // OK

Why is -1 > sizeof(int) true?

1. Is it true that -1 is promoted to unsigned(-1) and then unsigned(-1) > sizeof(int).
2. Is it true that -1 > sizeof(int) is equivalent to -1 > size_t(4) if sizeof(int) is 4. If this is so why -1 > size_t(4) is false?

Is this C++ standard comformant?

回答1:

The following is how standard (ISO 14882) explains abort -1 > sizeof(int)

Relational operator `>' is defined in 5.9 (expr.rel/2)

The usual arithmetic conversions are performed on operands of arithmetic or enumeration type. ...

The usual arithmetic conversions is defined in 5 (expr/9)

... The pattern is called the usual arithmetic conversions, which are defined as following:

• If either operand is of type long double, ...
• Otherwise, if either operand is dobule, ...
• Otherwise, if either operand is float, ...
• Otherwise, the integral promotions shall be performed on both operands.
• ...

The integral promotions is defined in 4.5 (conv.prom/1)

An rvalue of type char, signed char, unsigned char, short int, or unsigned short int can be converted to an rvalue of type int if int can represent all the values of the source type; otherwise, the source rvalue can be converted to an rvalue of type unsigned int.

The result of sizeof is defined in 5.3.3 (expr.sizeof/6)

The result is a constant of type size_t

size_t is defined in C standard (ISO 9899), which is unsigned integer type.

So for -1 > sizeof(int), the > triggers usual arithmetic conversions. The usual arithmetic conversion converts -1 to unsigned int because int cannot represent all the value of size_t. -1 becomes a very large number depend on platform. So -1 > sizeof(int) is true.

回答2:

Because unsigned is stronger then signed and -1 converted to unsigned value as of size_t , so actually -1 == 0xFFFFFFFF > 4

This is how it should work according to C++ standard

回答3:

because -1 gets casted to size_t and this is an unsigned data type - so (size_t)-1 == 4294967295 (on a 32bit system) which is definitely larger than 4

if you add -Wall to the gcc settings for example you get a warning that you are comparing a signed and an unsigned data type

回答4:

It's simple and sad. In C/C++:

1. most of the time, unsigned integer types have the semantic of modular integers (they represent equivalence classes)
2. comparisons of unsigned integer types have the semantic of usual integer ordering, so that 1U < 2U (IOW 0U is the smallest unsigned value)
3. sizeof has type size_t
4. size_t is an unsigned integer type
5. Point (1) implies that mixed arithmetic computations involving a signed and an unsigned integer are done in unsigned, modular arithmetic: this is the only possibility without violating "unsigned mean modular" rule. It's trivial to convert an integer to the equivalence class of integers equivalent to it. (Whereas going the other way requires the choice of an integer to represent the equivalence class.)
6. Point (5) implies that -1 < 1U is interpreted as unsigned(-1) < 1U, and unsigned(-1) = - 1U, and obviously - 1U < 1U, so -1 < 1U is true.
7. Points (1,3,4) imply that sizeof something acts (mostly) as an equivalent class (!!!).
8. All this imply that -1 < sizeof something

The conclusion: this is a design error inherited from C.

Rule:

Only use unsigned types for modular arithmetic, bits manipulations (&, |, ^, <<, >>, ~ operators), byte manipulations (unsigned char means "byte" in C/C++), and characters (unsigned char means character in C/C++).

Do not use unsigned types to do arithmetic.

If a function expects an integer value that should never be negative, take a signed integer, and optionally check in the function that the value is in range.