问题描述
可能的重复:
指针变量和指针变量有什么区别C++中的引用变量?
路过有好处吗在 C++ 中通过引用传递指针?
在这两种情况下，我都达到了结果.那么什么时候一个比另一个更受欢迎呢?我们使用一个而不是另一个的原因是什么?
In both cases, I achieved the result. So when is one preferred over the other? What are the reasons we use one over the other?
#include <iostream>
using namespace std;
void swap(int* x, int* y)
{
int z = *x;
*x=*y;
*y=z;
}
void swap(int& x, int& y)
{
int z = x;
x=y;
y=z;
}
int main()
{
int a = 45;
int b = 35;
cout<<"Before Swap
";
cout<<"a="<<a<<" b="<<b<<"
";
swap(&a,&b);
cout<<"After Swap with pass by pointer
";
cout<<"a="<<a<<" b="<<b<<"
";
swap(a,b);
cout<<"After Swap with pass by reference
";
cout<<"a="<<a<<" b="<<b<<"
";
}
输出
Before Swap
a=45 b=35
After Swap with pass by pointer
a=35 b=45
After Swap with pass by reference
a=45 b=35
推荐答案
引用在语义上如下:
T&<=>*(T * const)
const T&<=>*(T const * const)
T&&<=>[无 C 等价物] (C++11)
与其他答案一样，C++ 常见问题解答中的以下内容是单行答案:可能时引用，需要时使用指针.
As with other answers, the following from the C++ FAQ is the one-line answer: references when possible, pointers when needed.
优于指针的一个优点是您需要显式转换才能传递 NULL.不过还是有可能的.在我测试过的编译器中，没有一个会发出以下警告:
An advantage over pointers is that you need explicit casting in order to pass NULL. It's still possible, though. Of the compilers I've tested, none emit a warning for the following:
int* p() {
return 0;
}
void x(int& y) {
y = 1;
}
int main() {
x(*p());
}
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