c++ - 两个无符号相减得到有符号

Question

我有以下代码:

#include <cstdint>

template <typename T>
T test(T a, T b)
{
    float aabb = reinterpret_cast<float>(a - b);
}

int main(int argc, const char *argv[])
{
    std::uint8_t a8, b8;
    test(a8, b8);
    return 0;
}

我知道 reinterpret_cast<float>无法工作，并且在编译时会出错。我正在使用该错误，以便编译器告诉我 a - b 的类型.

问题是在这种情况下，它表示 a - b 的类型是int当它们都是 uint8_t (unsigned char) 时. uint16_t 也是如此.但不是 uint32_t它说 a - b是unsigned int .

所以，我的问题是:这是预期的行为(unsigned char - unsigned char 给出一个 int)，还是某种奇怪的编译器错误(同时使用 GCC 和 clang 进行测试)？

Answer 1

是的，这是预期的，作为所谓的常规算术转换的一部分，结合了积分提升的规则。

C++03 和 C++11 之间的确切措辞发生了变化，但在这种情况下最终结果是相同的。

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[C++03: 4.5/1]: 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.

[C++03: 4.5/5]: These conversions are called integral promotions.

[C++03: 5/9]: Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result.

This pattern is called the usual arithmetic conversions, which are defined as follows:

• If either operand is of type long double, the other shall be converted to long double.
• Otherwise, if either operand is double, the other shall be converted to double.
• Otherwise, if either operand is float, the other shall be converted to float.
• Otherwise, the integral promotions (4.5) shall be performed on both operands.⁵⁴
• Then, if either operand is unsigned long the other shall be converted to unsigned long.
• Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int.
• Otherwise, if either operand is long, the other shall be converted to long.
• Otherwise, if either operand is unsigned, the other shall be converted to unsigned.

[Note: otherwise, the only remaining case is that both operands are int ]

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[C++11: 4.5/1]: A prvalue of an integer type other than bool, char16_t, char32_t, or wchar_t whose integer conversion rank (4.13) is less than the rank of int can be converted to a prvalue of type int if int can represent all the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned int.

[C++11: 4.5/7]: These conversions are called integral promotions.

[C++11: 5.9]: Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result.

This pattern is called the usual arithmetic conversions, which are defined as follows:

• If either operand is of scoped enumeration type (7.2), no conversions are performed; if the other operand does not have the same type, the expression is ill-formed.
• If either operand is of type long double, the other shall be converted to long double.
• Otherwise, if either operand is double, the other shall be converted to double.
• Otherwise, if either operand is float, the other shall be converted to float.
• Otherwise, the integral promotions (4.5) shall be performed on both operands.⁵⁹ Then the following rules shall be applied to the promoted operands:
  • If both operands have the same type, no further conversion is needed.
  • Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank shall be converted to the type of the operand with greater rank.
  • Otherwise, if the operand that has unsigned integer type has rank greater than or equal to the rank of the type of the other operand, the operand with signed integer type shall be converted to the type of the operand with unsigned integer type.
  • Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, the operand with unsigned integer type shall be converted to the type of the operand with signed integer type.
  • Otherwise, both operands shall be converted to the unsigned integer type corresponding to the type of the operand with signed integer type.