通过指针&通过引用 [英] Pass by pointer & Pass by reference

问题描述

可能重复：



在C ++中，通过引用传递指针可以获得好处？

在这两种情况下，我都取得了结果。
那么什么时候优先于另一个呢？我们使用其中一个的原因是什么？

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\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

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] （C ++ 11）

与其他答案一样，C ++ FAQ中的以下是单行答案

As with other answers, the following from the C++ FAQ is the one-line answer: references when possible, pointers when needed.

指针的优点是，你需要显式转换以传递NULL。
仍然可以，但。
在我测试的编译器中，没有发出以下警告：

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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