c++ - 从有符号到无符号问题的类型转换

Question

我有一个类似下面的函数:

template<typename T>
void myF(void* a, T* b, T c) 
{
    make_unsigned<T>::type newC;
    make_unsigned<T>::type* newB = ptr_cast<make_unsigned<T>::type*>(b);
    ...
}

template <typename T> T ptr_cast(void* ptr)
{
    return static_cast<T>(ptr);
}

正在使用 type_traits 类。它在 VS 2010 中工作得很好，但是当我使用 ARM 编译器编译时它失败了。编译器是 v.5.02: http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.subset.swdev.ds5/index.html

我得到:找不到文件 type_traits...我认为此时它是标准库的一部分？

我尝试了 make_unsigned 的自定义实现:

namespace internal {

    #define MK_MAKEUNSIGNED(T,V)             \
    template<> struct make_unsigned<T> {     \
      public:                              \
        typedef V type;                    \
    };

    template<typename T>
    struct make_unsigned {
        typedef T type;
    };

    MK_MAKEUNSIGNED(sdata8, data8);
    MK_MAKEUNSIGNED(sdata16, data16);
    MK_MAKEUNSIGNED(sdata32, data32);
    MK_MAKEUNSIGNED(sdata64, data64);
    #undef MK_MAKEUNSIGNED

};

并修改为:

template<typename T>
void myF(void* a, T* b, T c) 
{
    internal::make_unsigned<T>::type newC;
    internal::make_unsigned<T>::type* newB = ptr_cast<internal::make_unsigned<T>::type*>(b);
    ...
}

同样，它在 VS 2010 中工作，但 ARM 编译器给出以下错误:

internal::make_unsigned<T>::type newC; #276 name followed by "::" must be a class or namespace name
^  
internal::make_unsigned<T>::type newC; #282 the global scope has no "type"
                            ^ 
internal::make_unsigned<T>::type newC; #65: expected a ';'
                                 ^
internal::make_unsigned<T>::type* newB = static_ptr<internal::make_unsigned<T>::type*>(b); #276 name followed by "::" must be a class or namespace name
^
internal::make_unsigned<T>::type* newB = static_ptr<internal::make_unsigned<T>::type*>(b); #282 the global scope has no "type" 
                            ^
internal::make_unsigned<T>::type* newB = static_ptr<internal::make_unsigned<T>::type*>(b); #20 identifier 'newB' nis undefined
                                  ^
internal::make_unsigned<T>::type* newB = static_ptr<internal::make_unsigned<T>::type*>(b); #276 name followed by "::" must be a class or namespace name
                                                    ^
internal::make_unsigned<T>::type* newB = static_ptr<internal::make_unsigned<T>::type*>(b); #276 name followed by "::" must be a class or namespace name
                                         ^
internal::make_unsigned<T>::type* newB = static_ptr<internal::make_unsigned<T>::type*>(b); #29 expected an expression                                                          
                                                                                     ^

所以我似乎无法使用 type_traits 或自定义实现来编译它。非常感谢任何想法!

Answer 1

您缺少两个 typename在这里和那里......

typename internal::make_unsigned<T>::type newC;
// ^^^^^

基本上，internal::make_unsigned<T>::type是从属名称，除非您使用 typename 指示编译器，否则假定它不是类型. VS 过于宽松，让我们过去吧。

除此之外，这不会编译:

typename make_unsigned<T>::type* newB = static_cast<make_unsigned<T>::type*>(b);

因为你不能static_cast从有符号指针到无符号指针。