C ++。为什么我不能编译这段代码？使用const_cast删除const const有什么问题？ [英] C++. Why I can't compile this code? What is wrong with removing constness using const_cast?

问题描述

我有一些问题，使用const_cast删除常量。错误msg说转换是一个有效的标准转换.....

I have some problem removing constness using const_cast. Error msg says "Conversion is a valid standard conversion....."

这个问题的性质是什么？为什么要使用C风格的转换？

What is the nature of this problem? Why should I use C-style cast instead?

错误C2440：'const_cast'：无法从'const size_t'转换是一种有效的标准转换，可以隐式执行，或者使用static_cast，C风格的转换或函数式转换

"error C2440: 'const_cast' : cannot convert from 'const size_t' to 'size_t'" "Conversion is a valid standard conversion, which can be performed implicitly or by use of static_cast, C-style cast or function-style cast"

template<typename T>
const IFixedMemory* FixedMemoryPkt<T>::operator=(const IFixedMemory* srcObj)
{
    // doesn't remove constness this way. why?
    const_cast<char*> (this->m_Address) = (char*)srcObj->GetAddress();

    // compile this way, but maybe(?) undefined behaviour
    // const_cast<char*&> (this->m_Address) = (char*)srcObj->GetAddress();

    // doesn't doesn't work too
    const_cast<size_t> (this->m_Size) = (size_t)(srcObj->GetSize());
    // const_cast<size_t> (this->m_Size) = 0;

    return this;
}

template<typename T>
class FixedMemoryPkt : public IFixedMemory
{
private:
    const size_t m_Size;
    const char*  m_Address;
}

class IFixedMemory
{
public:
    virtual const char* GetAddress() const = 0;
    virtual size_t GetSize() const = 0;
}

推荐答案

const_cast 用于将指针或对 const 对象的引用转换为非 - const 等同物。但是，如果对象本身 const ，则不能使用它们修改它们引用的对象。没有有效的方法来修改 m_Size ;如果你想修改它，那么不要声明它 const 。

const_cast is used to convert from pointers or references to const objects, to their non-const equivalents. However, you can't use them to modify the object they refer to if the object itself is const. There is no valid way to modify m_Size; if you want to modify it, then don't declare it const.

分配给指针，因为指针本身不是 const ：

You do not need a cast to assign to the pointer, since the pointer itself is not const:

this->m_Memory = srcObj->GetAddress();

如果您确实希望指针本身为 const ，则 * ：

If you did want the pointer itself to be const, then the const would come after the *:

char * const m_Address;

如错误所示，您可以转换 const > value 转换为未经转换的值的非 - const 临时副本但您无法分配到该临时。

As the error says, you can convert a const value into a non-const temporary copy of that value without a cast; but you couldn't assign to that temporary.

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