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      2. 为什么没有从指针到引用到 const 指针的隐式转换

        时间:2023-12-02

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                1. 本文介绍了为什么没有从指针到引用到 const 指针的隐式转换的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  我将用代码说明我的问题:

                  I'll illustrate my question with code:

                  #include <iostream>
                  
                  void PrintInt(const unsigned char*& ptr)
                  {
                      int data = 0;
                      ::memcpy(&data, ptr, sizeof(data));
                      // advance the pointer reference.
                      ptr += sizeof(data);
                      std::cout << std::hex << data << " " << std::endl;
                  }
                  
                  int main(int, char**)
                  {
                      unsigned char buffer[] = { 0x11, 0x11, 0x11, 0x11, 0x22, 0x22, 0x22, 0x22, };
                  
                      /* const */ unsigned char* ptr = buffer;
                  
                      PrintInt(ptr);  // error C2664: ...
                      PrintInt(ptr);  // error C2664: ...    
                  
                      return 0;
                  }
                  

                  当我运行这段代码(在 VS2008 中)时,我得到这个:错误 C2664:'PrintInt':无法将参数 1 从 'unsigned char *' 转换为 'const unsigned char *&'.如果我取消注释const"注释,它就可以正常工作.

                  When I run this code (in VS2008) I get this: error C2664: 'PrintInt' : cannot convert parameter 1 from 'unsigned char *' to 'const unsigned char *&'. If I uncomment the "const" comment it works fine.

                  但是指针不应该隐式转换为const指针然后引用吗?我期望这行得通有错吗?谢谢!

                  However shouldn't pointer implicitly convert into const pointer and then reference be taken? Am I wrong in expecting this to work? Thanks!

                  推荐答案

                  如果指针被转换为 const 指针,如您所建议的,那么转换的结果是一个临时值,一个 rvalue.您不能将非常量引用附加到右值 - 这在 C++ 中是非法的.

                  If the pointer gets converted to a const pointer, as you suggest, then the result of that conversion is a temporary value, an rvalue. You cannot attach a non-const reference to an rvalue - it is illegal in C++.

                  例如,由于类似的原因,此代码将无法编译

                  For example, this code will not compile for a similar reason

                  int i = 42;
                  double &r = i;
                  

                  即使 int 类型可以转换为 double 类型,它仍然不意味着你可以附加一个 double & 引用到该转换的结果.

                  Even though type int is convertible to type double, it still doesn't mean that you can attach a double & reference to the result of that conversion.

                  然而,const 引用(即引用到const 类型的引用)可以附加到右值,这意味着这段代码可以完美地编译

                  However, a const reference (i.e. a reference of reference-to-const type) can be attached to an rvalue, meaning that this code will compile perfectly fine

                  int i = 42;
                  const double &r = i;
                  

                  在您的情况下,如果您将函数声明为

                  In your case if you declare your function as

                  void PrintInt(const unsigned char* const& ptr) // note the extra `const`
                  

                  代码将被编译.

                  这篇关于为什么没有从指针到引用到 const 指针的隐式转换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跟版网!

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