Question

I am trying to understand a program, which includes the following definition for a function f

void f(String S, const String& r)
{

}

Here String in the argument stands for a class. I am confused on the difference between the definitions of these two arguments: “String S” and “const String& r”. S should represent an object of class String, then how about r?

In more detail, the f is defined as

void f(String S, const String& r)
{
   int c1 = S[1];  // c1=s.operator[](1).operator char( )
   s[1] ='c';      // s.operator[](1).operator=('c')

   int c2 = r[1];  // c2 = r.operator[](1)
   r[1] = 'd';     // error: assignment to char, r.operator[](1) = 'd'      
 }

This code snippet is to show how the operator overload, but these comments does not help me much. For instance, why r[1]=’d’ is nor correct? Thanks for helping understanding it.

Answer 1

const String& r is a constant reference to String r. Within the function, you access r just like a String. The difference is that it is actually a reference to the object passed to the function, while S will be a copy of the object passed to the function. You can almost think of it as if you are accessing r through a de-referenced pointer (though there is more to it than that).

Another way to look at it: The caller will see changes to r (if it wasn’t const), while he will not see changes to S.

The const simply means the function f cannot modify r.

See also: https://isocpp.org/wiki/faq/references#overview-refs