By value or by reference?

[Jacek Generowicz]

aleaxit at yahoo.com (Alex Martelli) writes:

> Jacek Generowicz <jacek.generowicz at cern.ch> wrote:
>    ...
> > Here's a program that tells you the answer. Should be trivially
> > translatable to many other languages.
> > 
> > ========================================================================
> > def passCheck(b):
> >     b = 'new'
> > 
> > def doesThisLanguageBindByValue(language):
> >     a = 'original'
> >     b = a
> >     b = 'new'
> >     if a == 'original':
> >         print "%s assigns by value" % (language)
> >     else:
> >         print "%s does not assigns by value" % (language)
> 
> Can you give some examples of languages which would end up in the else
> branch?

Sure. Here it is in C++:

========================================================================
#include <iostream>
#include <string>

void passCheck(std::string& b) {
  b = "new";
}

void doesThisBindByValue() {
  std::string  a = "original";
  std::string& b = a;
  b = "new";
  if (a == "original") {
    std::cout << "by value" << std::endl;
  } else {
    std::cout << "NOT by value" << std::endl;
  }

  a = 1;
  passCheck(a);
  if (a == "original") {
    std::cout << "by value" << std::endl;
  } else {
    std::cout << "NOT by value" << std::endl;
  }
}

int main() {

  doesThisBindByValue();

  return 0;
}
========================================================================

> I think that with a suitably perverse operator= I might be able
> to imitate this weird effect in C++, but then I think I'd have to make
> the 'b = a' a copy ctor instead of an assignment (or get to such levels
> of perversion that even my once-C++-addled mind reels at...;-).

No need for such horrors. Just declare the variables to be references
... and then you really do get bind-by-reference. It does exactly what
is says on the tin (unusual for C++, I know :-).