无符号短减法的类型

Question

我有一个简短的程序，不编译：

    #include <algorithm>

    int main(int argc, char *argv[])
    {
        unsigned short a = 490;
        unsigned short b = 43;

        unsigned short d = std::min(b, a-b);

        return 0;
    }

编译器（g ++版本4.8.4）说：

    ./main.cpp: In function ‘int main(int, char**)’:
    ./main.cpp:25:47: error: no matching function for call to ‘min(short unsigned int&, int)’
                 unsigned short d = std::min(b, a-b);
                                                   ^
    ./main.cpp:25:47: note: candidates are:
    In file included from /usr/include/c++/4.8/algorithm:61:0,
                     from ./main.cpp:18:
    /usr/include/c++/4.8/bits/stl_algobase.h:193:5: note: template<class _Tp> const _Tp& std::min(const _Tp&, const _Tp&)
         min(const _Tp& __a, const _Tp& __b)
         ^
    /usr/include/c++/4.8/bits/stl_algobase.h:193:5: note:   template argument deduction/substitution failed:
    ./main.cpp:25:47: note:   deduced conflicting types for parameter ‘const _Tp’ (‘short unsigned int’ and ‘int’)
                 unsigned short d = std::min(b, a-b);
                                                   ^
    In file included from /usr/include/c++/4.8/algorithm:61:0,
                     from ./main.cpp:18:
    /usr/include/c++/4.8/bits/stl_algobase.h:239:5: note: template<class _Tp, class _Compare> const _Tp& std::min(const _Tp&, const _Tp&, _Compare)
         min(const _Tp& __a, const _Tp& __b, _Compare __comp)
         ^
    /usr/include/c++/4.8/bits/stl_algobase.h:239:5: note:   template argument deduction/substitution failed:
    ./main.cpp:25:47: note:   deduced conflicting types for parameter ‘const _Tp’ (‘short unsigned int’ and ‘int’)
                 unsigned short d = std::min(b, a-b);

所以显然无符号短减法产生一个整数。

问题是：为什么？

编辑： 我在搜索时看到了副本，但没有读到最后。他们说： 的＆＃34;注意。最小操作大小为int。因此在操作完成之前将short / char提升为int。＆＃34;这解释了行为......