Re: lambda capture using references in scope

From:
TheGunslinger <mikiesrunsbaal.sec@sbcglobal.net>
Newsgroups:
comp.lang.c++.moderated
Date:
Wed, 1 Aug 2012 21:27:43 -0700 (PDT)
Message-ID:
<qmjj18ltqj1krh4alfl2d0k8e6hf291g30@4ax.com>
On Wed, 25 Jul 2012 10:03:26 -0700 (PDT), "gast128@hotmail.com"
<gast128@hotmail.com> wrote:

Hello all,

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);
  });
}

Using vs2010.


Here is a code snippet from Sams Learn C++ in One Hour a Day, 7th
Edition, which covers Lambda functions (and the C++11 standard
updates) and expressions, page 161, Listing 7.11:

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

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