在C ++中为对象分配引用的值时会发生什么？

Question

#include <iostream>

void test(std::string &s) {
    std::string p = s;
    std::cout << p.length() << " " << p;
}

int main() {
   std::string s = "Hello world";
   test(s);

   return 0;
}

So, the function test receives a reference to my string s from my main function.

My question is, what does this line do:

std::string p = s;

Does it shallow copy the reference and put it in p , thereby defeating the purpose of using a reference in the first place?

Or does it ( p ) just act as a reference?

Answer 1

它创建引用的值的副本，不允许您使用p编辑s的内容，以使p充当引用并能够从p编辑s ，您必须将其声明为引用：

std::string& p = s;

Answer 2

thereby defeating the purpose of using a reference in the first place?

Why does it defeat the purpose of reference, you have declared s as reference, isn't it? Not p . So as noted indeed it will copy the value which s contains.

Answer 3

What happens when I assign an object the value of a reference in C++?

You cannot assign the value of a reference, or at least you should not think in such terms. A reference is not a pointer. What happens is that you assign the value of an object, because a reference is the object , accessed via a different name.

 void test(std::string &s) { 

The fact that you are dealing with a reference is only really relevant at the point of declaration. All code in this function which uses s uses a std::string , not a std::string & .

My question is, what does this line do:

 std::string p = s; 

It assigns your std::string object to p , no more, no less. In other words, it does the same as it would in this program:

int main() {
   std::string s = "Hello world";

   std::string &s_with_different_name = s;
   std::string p = s_with_different_name;
   std::cout << p.length() << " " << p;
}

Or, even more simply, in this:

int main() {
   std::string s = "Hello world";

   std::string p = s;
   std::cout << p.length() << " " << p;
}