i was surprised the other day that when u pass references in lambda's they actually get copied when capturing by value. If u pass a pointer, the pointer gets copied, but not the pointee. However if u pass a reference also the thing the reference points to
get copied. Its the reverse what I would expect (atm) but oth the c++ committee often does that :(. But maybe someone has a powerful explanation?

Example:

void Foo()
{
   std::string       str;
   std::string&      rstr = str;
   std::vector<int>  vec(1);

   printf("%08x", &rstr);

   //a copy of rstr will be made???
   std::for_each(vec.cbegin(), vec.cend(), [=] (int n)
   {
      //rstr gets copied
      printf("%08x", &rstr);
   });

   std::for_each(vec.cbegin(), vec.cend(), [&] (int n)
   {
      printf("%08x", &rstr);
   });

void DisplayNums(vector<int>& DynArray)
        {
        for_each (DynArray.begin(), DynArray.end(),  \
      [] (int Element) {cout << Element << " ";} ); // Lambda

        cout << endl;
        }

Since vectors are a template class, you should be able to substitute
<str> or <char> for <int>

Hope this helps.

Respectfully,

MJR

 John
--
John Harris

Since your first lambda (anonymous function) specifies 'capture
(i.e. pass) by value' the lambda function behaviour is consitent with
normal parameter passing in C++.

Hans

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Hi, not exactly sure I understand what you mean.

Your first lambda captures by value (i.e. [=]). So if you grab an entity
inside the lambda which is declared as a reference in the enclosing
environment you will get a copy of it since you implicitly have stated
that the environment should be passed by value into the lambda. This is
consistent with normal parameter passing.

I know that you asked for a reason for the behaviour (which I hope I
gave),not a ref to the standard, but here is the standard anyway:

------ 5.1.2 ------
14 An entity is captured by copy if it is implicitly captured and the
capture default is = or if it is explicitly captured with a capture that
does not include an &. For each entity captured by copy, an unnamed
nonstatic data member is declared in the closure type. Blah, blah ...

15 An entity is captured by reference if it is implicitly or explicitly
captured but not captured by copy. Blah, blah ...
------

In your case for your first lambda, point 14 is applicable.

Regards,
Hans