指针和引用参数之间的区别?-C/C++问题

difference between a pointer and reference parameter?(指针和引用参数之间的区别?)

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问题描述

这些都一样吗:

int foo(bar* p) {返回 p->someInt();}

和

int foo(bar& r) {返回 r.someInt();}

忽略空指针的潜力.无论 someInt() 是虚拟的还是传递一个 bar 或 bar 的子类，这两个函数在功能上是否相同?>

这个切片有什么作用:

bar&ref = *ptr_to_bar;

解决方案

C++ 引用在标准中有意未指定使用指针实现.引用更像是变量的同义词"，而不是指向它的指针.当有可能意识到指针在某些情况下是多余的时，这种语义为编译器打开了一些可能的优化.

还有一些区别:

您不能将 NULL 分配给引用.这是一个关键的区别，你更喜欢一个的主要原因其他.
当你获取一个地址时指针，你得到的地址指针变量.当你拿参考地址，您将获得变量的地址是提及.
您无法重新分配参考.初始化后，它在整个生命周期中都指向同一个对象.

Are these the same:

int foo(bar* p) {
  return p->someInt();
}

and

int foo(bar& r) {
  return r.someInt();
}

Ignore the null pointer potential. Are these two functions functionally identical no matter if someInt() is virtual or if they are passed a bar or a subclass of bar?

Does this slice anything:

bar& ref = *ptr_to_bar;

解决方案

C++ references are intentionally not specified in the standard to be implemented using pointers. A reference is more like a "synonym" to a variable than a pointer to it. This semantics opens some possible optimizations for the compiler when it's possible to realize that a pointer would be an overkill in some situations.

A few more differences:

You can't assign NULL to a reference. This is a crucial difference and the main reason you'd prefer one over the other.
When you take the address of a pointer, you get the address of the pointer variable. When you take the address of a reference, you get the address of the variable being referred to.
You can't reassign a reference. Once it is initialized it points to the same object for its entire life.

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