Question

Why does the following code compile with a circular const variable referencing itself?

#include <iostream>

extern int main(int argc, char* argv[])
{
  const int foo = foo * 60;
  std::cout << foo << std::endl;
  return 0;
}

I’m compiling on a Solaris 5.10 x86 host with the SUNWspro compiler:

/opt/SUNWspro/bin/CC test.cpp

For completeness, this is what it prints:

$ ./a.out
-519270512

Answer 1

In C++, variables are in scope and can be used as part of their own initializers. For example, the following code is also legal:

int x = x;

Doing this results in undefined behavior, since you’re referring to the value of x before it has been initialized. However, it’s perfectly legal code. I think that the rationale behind this is that you might in some cases want an object to refer to itself during its construction. For example, you could conceivably do something like this:

MyObject x(137, &x); // Pass a pointer to x into its own constructor

I’m honestly not sure why you’d ever want to do this, though.

Hope this helps!