C ++“转换失去限定符”编译错误 [英] C++ "conversion loses qualifiers" compile error

问题描述

我在今天调试SWIG类型的电子书时遇到了一个有趣的问题。任何人都关心我为什么Visual C ++ 2008在转换从 myLib :: Char * 到时抛出转换失去限定符错误const ourLib :: Char *& ？我认为类型* - > const类型* 是一个平凡的转换， c> Lvalue - > Lvalue& 。

I ran into an interesting problem while debugging SWIG typemaps today. Anyone care to enlighten me why Visual C++ 2008 throws a "conversion loses qualifiers" error when converting from ourLib::Char * to const ourLib::Char * &? I thought Type * -> const Type * was a trivial conversion, and (when calling functions) Lvalue -> Lvalue & as well.

我们结束了：

// ourLib::Char is a typedef'ed char on Win32

%typemap(in) const char* (const ourLib::Char* tmp)
{
    if (!bapiLua::LuaTraits<ourLib::Char*>::FromLuaObject(L, $argnum, tmp)) SWIG_fail;
    $1 = const_cast<char *>(tmp);
}

// And in a different source file, already written:
namespace bapiLua {
template<>
struct LuaTraits<ourLib::Char*>
{
    static ourLib::Bool FromLuaObject(lua_State* L, int pos, const ourLib::Char*& o_result);
};
}

删除 const 从 const ourLib :: Char * tmp 导致我描述的错误。

Removing the const from const ourLib::Char * tmp causes the error I described.

推荐答案

p>说你有以下函数：

Say you had the following function:

void test(  const char*& pRef)
{
    static const char somedata[] = { 'a' ,'b', 'c', '\0'};
    pRef = somedata;
}

如果你传递一个非const ，那么当 test（）返回时，编译器将丢失以下事实： p 指向的是 const 。

If you passed in a non-const char*, then when test() returned the compiler would have lost the fact that what p is pointing to is const.

这本质上与在这个C ++ FAQ Lite问题

It's essentially the same reason as given in this C++ FAQ Lite question (dealing with pointers-to-pointers rather than pointer references):

• http://www.parashift.com/c++-faq-lite/const-correctness.html#faq-18.17

• http://www.parashift.com/c++-faq-lite/const-correctness.html#faq-18.17

这篇关于C ++“转换失去限定符”编译错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！