C ++ const_cast用法而不是C风格的转换 [英] C++ const_cast usage instead of C-style casts

问题描述

为什么是以下内容：

  const int i0 = 5;
//int       i1 = const_cast<int>(i0);       // compilation error
  int       i2 = (int)i0;                   // okay

  int       i3 = 5;
//const int i4 = const_cast<const int>(i3); // compilation error
  const int i5 = (const int)i3;             // okay

推荐答案

  const int i0 = 5;
//int       i1 = const_cast<int>(i0);       // compilation error
  int       i2 = (int)i0;                   // okay

  int       i3 = 5;
//const int i4 = const_cast<const int>(i3); // compilation error
  const int i5 = (const int)i3;             // okay

编译错误是因为你不强制const / add const。相反，你复制i0。对于该操作，根本不需要转换：

The compilation errors are caused because you don't cast const away/add const. Instead, you copy i0. For that operation, no cast is required at all:

int i1 = i0;
const int i4 = i3;

你转换的类型实际上应该是一个指针或引用。否则，使用const_cast没有意义，因为你可以直接复制它。例如，对于指针，可以抛弃const，因为取消引用指针将为 const T * 产生另一个类型（产生 const T ） T * （产生 T ）。对于引用，同样是真的： T& 将使用另一个 this 指针类型访问对象，而不是使用 const T& 。现在你真正想要归档：

The type you cast to should actually be a pointer or reference. Otherwise, using const_cast doesn't make sense since you could straight copy it. For pointers, for example, you can cast away the const, because dereferencing the pointer will yield another type for a const T* (yields const T) than for a T* (yields T). For references, the same is true: T& will access the object using another this pointer type than using const T&. Now what you really wanted to archive:

  const int i0 = 5;
//int &     i1 = const_cast<int&>(i0);      // okay (but dangerous)
  int &     i2 = (int&)i0;                  // okay (but dangerous)

  int       i3 = 5;
//const int&i4 = const_cast<const int&>(i3); // ok now and valid!
  const int&i5 = (const int&)i3;             // okay too!

上面的例子可能导致未定义的行为，当你写一个原来的const对象通过引用非-const（实际上，只是转换和读取它本身不是未定义的行为，但如果你抛弃const，你也可以写它，然后产生未定义的行为）

The above can lead to undefined behavior, when you write to an originally const object through a reference to non-const (actually, merely casting and reading it isn't undefined behavior in itself. But if you're casting away const, you may also write to it, which then yields the undefined behavior)

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