c++ - 为什么 int 加上 uint 返回 uint？

Question

int 加上 unsigned int 返回一个 unsigned int。应该这样吗？

考虑这段代码:

#include <boost/static_assert.hpp>
#include <boost/typeof/typeof.hpp>
#include <boost/type_traits/is_same.hpp>

class test
{
    static const int          si = 0;
    static const unsigned int ui = 0;

    typedef BOOST_TYPEOF(si + ui) type;
    BOOST_STATIC_ASSERT( ( boost::is_same<type, int>::value ) ); // fails
};


int main()
{
    return 0;
}

Answer 1

如果“应该”是指“我的编译器的行为是否符合标准”:是。

C++2003:第 5 条第 9 段:

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:

• blah
• Otherwise, blah,
• Otherise, blah, ...
• Otherwise, if either operand is unsigned, the other shall be converted to unsigned.

如果“应该”是指“如果不这样做，世界会变得更好吗”:我没有资格回答这个问题。