当通过引用传递时取消引用指针 [英] dereferencing a pointer when passing by reference

问题描述

当通过引用一个函数来解引用指针时会发生什么？

这是一个简单的例子

  int& returnSame（int& example）{return example; } 
 
 int main（）
 {
 int inum = 3; 
 int * pinum =&英语
 
 std :: cout<< inum：< returnSame（* pinum）<< std :: endl; 
 
 return 0; 
 
} 
  

是否生成了临时对象？

解决方案

解除引用指针不会创建副本;它创建指向指针的目标的 lvalue 。这可以绑定到 lvalue 引用参数，因此函数接收对指针指向的对象的引用，并返回对该引用的引用。此行为是定义明确的，不涉及临时对象。

如果它接受的值是参数，那么会创建一个本地副本，并返回一个引用将是坏的，给定未定义的行为，如果它被访问。 / p>

what happens when you dereference a pointer when passing by reference to a function?

Here is a simple example

int& returnSame( int &example ) { return example; }

int main()
{
  int inum = 3;
  int *pinum = & inum;

  std::cout << "inum: " <<  returnSame(*pinum) << std::endl;

  return 0;          

}

Is there a temporary object produced?

解决方案

Dereferencing the pointer doesn't create a copy; it creates an lvalue that refers to the pointer's target. This can be bound to the lvalue reference argument, and so the function receives a reference to the object that the pointer points to, and returns a reference to the same. This behaviour is well-defined, and no temporary object is involved.

If it took the argument by value, then that would create a local copy, and returning a reference to that would be bad, giving undefined behaviour if it were accessed.

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