Why can a non-const member function be invoked on a constant structure?

Question

The program below compiles (with gcc), but should it? I would have thought that V1(1.0) created below is a constant, and so a non const method could not be invoked on it.

class V{
  double v;
 public:
  V(double v1){ v = v1;}
  void clear(){ v = 0;}
};

int main(){
  V(1.0).clear();
}

Compare this to a function "void f(int &t){}" which cannot be called as "f(1)", because 1 is a constant which cannot be a value for a non-const reference t.

I'm not clear how anything can be `const`. The keyword doesn't appear anywhere in your code. Also, member initialization syntax. — chris, Apr 24 '12 at 11:41
You also might want to look into r-value references (C++11) for your `void f(int &t){}` assumption. A bit different, but very similar concept. — chris, Apr 24 '12 at 11:44

score 3 · Answer 1 · answered Apr 24 '12 at 11:43

3

V(1.0) calls the constructor, which initializes double v1 by copy. So you have a temporary with a copy of the literal.

answered Apr 24 '12 at 11:43

juanchopanza

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score 1 · Answer 2 · answered Apr 24 '12 at 11:41

1

You seem to be confusing constants with r-values. f(1) would be illegal because 1 is not an l-value. V(1.0) is not a constant, but a temporary.

answered Apr 24 '12 at 11:41

Luchian Grigore

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I was supposing that temporaries should not appear as l-values. – Ekalavya Apr 24 '12 at 12:51

Why can a non-const member function be invoked on a constant structure?

2 Answers2