通过引用传递给构造函数 [英] Passing by reference to a constructor

问题描述

我决定看看给成员分配引用是否会使成员成为引用。我编写了以下代码片段进行测试。有一个简单的类 Wrapper ，其中一个 std :: string 作为成员变量。我在构造函数中使用 const string& 并将其分配给public成员变量。稍后在 main（）方法中，我修改了成员变量，但是我传递给构造函数的 string 保持不变，如何来？我认为在Java中变量会更改，为什么不在此代码段中？

I decided to see if assigning a reference to a member would make a member a reference. I wrote the following snippet to test it. There's a simple class Wrapper with an std::string as a member variable. I take take a const string& in the constructor and assign it to the public member variable. Later in the main() method I modify the member variable but the string I passed to the constructor remains unchanged, how come? I think in Java the variable would have changed, why not in this code snippet? How exactly do references work in this case?

#include <iostream>
#include <string>
using namespace std;

class Wrapper
{
public:
   string str;

   Wrapper(const string& newStr)
   {
      str = newStr;
   }
};

int main (int argc, char * const argv[]) 
{
   string str = "hello";
   cout << str << endl;
   Wrapper wrapper(str);
   wrapper.str[0] = 'j'; // should change 'hello' to 'jello'
   cout << str << endl;
}

推荐答案

在构造函数，您需要有一个引用成员

To assign a reference in a constructor you need to have a reference member

 class A{
     std::string& str;
 public:
     A(std::string& str_)
     :    str(str_) {} 
 };

str现在是对您传入的值的引用。const refs同样适用

str is now a reference to the value you passed in. Same applies for const refs

 class A{
     const std::string& str;
 public:
     A(const std::string& str_)
     :    str(str_) {} 
 };

但是，请记住，一旦分配了引用，就无法更改，因此如果需要分配更改str则必须改为使用指针。

However don't forget that once a reference has been assigned it can not be changed so if assignment requires a change to str then it will have to be a pointer instead.

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