通过引用传递时取消引用指针
通过引用传递给函数时取消引用指针会发生什么?
what happens when you dereference a pointer when passing by reference to a function?
这是一个简单的例子
int& returnSame( int &example ) { return example; }
int main()
{
int inum = 3;
int *pinum = & inum;
std::cout << "inum: " << returnSame(*pinum) << std::endl;
return 0;
}
是否产生了临时对象?
推荐答案
取消引用指针不会创建副本；它创建一个 lvalue 指向指针的目标.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，并返回对相同对象的引用.这种行为是明确定义的，不涉及临时对象.
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.
如果它按值获取参数，那么这将创建一个本地副本，并返回对该副本的引用会很糟糕，如果访问它会产生未定义的行为.
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.
相关文章