From: daniel.kruegler@googlemail.com   
      
   Am 10.04.2012 23:38, schrieb Martin B.:   
   > On 10.04.2012 20:00, Ulrich Eckhardt wrote:   
   [..]   
   >> [..] Also, any function can cheat and const_cast, so a   
   >> calling function must not assume that the call doesn't modify the   
   >> passed object. ...   
   >   
   > Interesting point. Who at which level is required to assume that   
   > something passed by const-ref is modified by the callee?   
      
   Any object that is not a const object may be modified. In most daily   
   scenarios arguments provided to functions are not really const objects,   
   therefore a "const &T" does not really protect from modification. Worse,   
   in the current standard it is not even clear whether an object with   
   dynamic storage duration (e.g. created by new expressions) can be a   
   const object. The intention is that they can be, see   
      
   http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#1428   
      
   Back to your last question: If you define a const object and provide   
   this object to any function (by reference or pointer), you can rely on   
   the assumption that code that would attempt to modify the object would   
   enter undefined behaviour. But as Ulrichs response implied: Compilers   
   would have a hard time to track any possible modifications of objects   
   that are referenced as const objects. For this reason they usually do   
   not optimize away objects provided by reference.   
      
   HTH & Greetings from Bremen,   
      
   Daniel Krügler   
      
      
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