c++ - 指针传递和引用传递

Question

Possible Duplicate:
What are the differences between pointer variable and reference variable in C++?
Are there benefits of passing by pointer over passing by reference in C++?

在这两种情况下，我都取得了结果。 那么什么时候比另一个更受欢迎呢？我们使用其中一种的原因是什么？

#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\n";
    cout<<"a="<<a<<" b="<<b<<"\n";

    swap(&a,&b);
    cout<<"After Swap with pass by pointer\n";
    cout<<"a="<<a<<" b="<<b<<"\n";

    swap(a,b);
    cout<<"After Swap with pass by reference\n";
    cout<<"a="<<a<<" b="<<b<<"\n";
}

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

Answer 1

引用在语义上如下:

T& <=> *(T * const)

const T& <=> *(T const * const)

T&& <=> [no C equivalent] (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:尽可能引用，需要时提供指针。

优于指针的一个优点是您需要显式转换才能传递 NULL。 不过，这仍然是可能的。 在我测试过的编译器中，没有一个会发出以下警告:

int* p() {
    return 0;
}
void x(int& y) {
  y = 1;
}
int main() {
   x(*p());
}