Passing by reference to a constructor

Question

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;
}

Answer 1

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 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_) {} 
 };

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.

Answer 2

Because Wrapper::str is not a reference, it's an independent object. So when you do str = newStr , you're copying the string.

Answer 3

class Wrapper
{
public:
   string& str;

   Wrapper(string& newStr) : str(newStr) {}
};

Note, you cannot accept a const string& and store it in a string& , you would lose const-correctness in doing so.

Answer 4

You need to use an initializer and declare str as a reference as in:

class Wrapper {
public:
   string &str;

   Wrapper(string& newStr)
      : str(newStr) {
   }
};

The way you're writing it, all you are doing is copying the value of the reference you pass to the constuctor. You're not saving the reference. By declaring a reference as a class member and initializing it with a reference to another string instance, you will get the behavior you're looking for.

Answer 5

Your main body variable is std::string .

Your parameter variable is const std::string& .

The const in references are always "low level const", meaning it modifies the type of object not the actual object.

In contrast a "top level const" modifies an actual object. Read C++ Primer on Top level const for clarification.

Here is how your assignment looks like when you pass arguments:

const std::string& = std::str; //Values are ommited
// i.e const std::string newStr = std::string str

You are initializing a const type reference with a non-const value which is acceptable. You are not supposed to change the value of std::string str using that reference. And, if you try changing the value of newStr inside the constructor, you will get a compilation error .

Next you're doing another assignment inside the constructor which is also acceptable:

std::string = const std::string 

The fact that wrap.str[0] didn't change str of main is that, although a reference was used to instantiate class str , class str has its own object and is not linked to main str . Using that reference in a parameter just links that parameter to main str ; not main str to class str .

If your class variables were referenced, then it could have changed.

Answer 6

您应该将Wrapper::str声明为string&而不是string 。