Are rvalues always constant?

When int i = 5;

int j = i;

Will rvalue i in this expression be a constant when evaluating the result?

I'm asking this question because in my copy constructor its argument requires a co开发者_StackOverflow社区nst

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There is a common misunderstanding around the lvalue/rvalue terms. They do not refer to variables, but rather to expressions. An expression can yield either an lvalue or an rvalue, and that can be either const or non-const.

In particular, in your code the expression i on the right hand side of the definition int j = i; is an lvalue expression, not an rvalue. For the purpose of assignment there is an lvalue to rvalue conversion and then that is assigned to the newly declared variable.

Cont-ness is an orthogonal concept --in most cases-- and relates to whether you can or cannot mutate the object that you are dealing with.

int f();
int& g();
const int& h();
const int k();

int main() {
  f();         // non-const rvalue expression
  g();         // non-const lvalue expression
  h();         // const lvalue expression
  k();         // const rvalue expression
  f() = 5;     // error, cannot assign to an rvalue
  g() = 5;     // correct, can modify a non-const lvalue
  h() = 5;     // error, cannot modify a constant lvalue
}

Other examples require the use of user defined types:

struct test {
   void foo() { x = 5; }
   void bar() const;
   int x;
};
test f();
const test g();
int main() {
   f().foo();      // f() is a non-const rvalue, 
                   // but you can call a method on the resulting object
   g().foo();      // g() is a const rvalue, 
                   // you cannot call a mutating member function
   g().bar();      // but you can call a const member function
}

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In C++, rvalues of built-in type cannot be const or non-const. It just doesn't make sense. There can be, however, const and non-const rvalues of class types.

An rvalue is just the VALUE (not the object/variable). What would you understand with "non-constant value" ?!