Quick Q: How does return by rvalue reference work? -- StackOverflow

By Blog Staff | May 8, 2013 01:27 PM | Tags: basics

Quick A: So easily that it's automatic... just return a value, not a reference.

How does return by rvalue reference work?

Just when I thought I kind of understand rvalue reference, I ran into this problem. The code is probably unnecessarily long, but the idea is quite simple. There is a main() function, and returnRValueRef() function.

[...]

AClass && returnRValueRef() {
  AClass a1(4);
  return move(a1);
}

int main() {
  AClass a;
  a = returnRValueRef();
}

Quick Q: What's the difference between vector<T...> and vector<T>...? -- StackOverflow

By Blog Staff | May 8, 2013 09:22 AM | Tags: intermediate

SO's version of the question added one layer of wrapping:

What's the difference between doing vector<vector<T...>> and vector<vector<T>...>

I saw code like this earlier:

 

using A = std::vector<std::vector<T>...>

where T is a variadic list of template arguments. I wanted to know what the difference is between putting the parameter pack at the end of the last angle bracket and the first. For example:

using B = std::vector<std::vector<T...>>;

Both of these two compile fine but I don't know what the difference is.

Can someone explain? Thanks.

Introduction To the C++11 Feature: Delegating Constructors -- sumi_cj

By Blog Staff | May 7, 2013 11:35 AM | Tags: basics

Head on over to the C/C++ Cafe for:

Introduction to the C++11 feature: delegating constructors

by sumi_cj

Excerpt:

In C++98, if a class has multiple constructors, these constructors usually perform identical initialization steps before executing individual operations. In the worst scenario, the identical initialization steps are copied and pasted in every constructor. See the following example: ...

Introduction To the C++11 Feature: Extended friend Declaration -- FangLu

By Blog Staff | May 6, 2013 11:26 PM | Tags: intermediate basics

Head on over to the C/C++ Cafe for:

Introduction to the C++11 feature: extended friend declaration

by FangLu

Excerpt:

The extended friend declaration feature is newly introduced in the C++11 standard. In this article, I will introduce this feature and provide some examples on how to use this feature.

 

Firstly, let's see why this feature is added into C++11. ...

GotW #1: Variable Initialization—or Is It? -- Herb Sutter

By Blog Staff | May 6, 2013 06:46 PM | Tags: intermediate basics

Herb Sutter is resuming his Guru of the Week series of problem-and-solution articles about coding in C++, with the intent to gradually update the 88 existing issues and write a few more along the way.

The first problem was posted today:

GotW #1: Variable Initialization -- or Is It? (3/10)

by Herb Sutter

This first problem highlights the importance of understanding what you write. Here we have a few simple lines of code — most of which mean something different from all the others, even though the syntax varies only slightly.

The solution is coming "soon"...

Ownership and 'Memory -- Andy Balaam

By Blog Staff | May 3, 2013 04:41 PM | Tags: basics

A short and basic summary of C++'s view of memory management: You can worry about it a lot less, and still be efficient, if you say who owns what.

Goodness in programming languages, part 4 -- Ownership & Memory

by Andy Balaam

From the post:

... over time the community of C++ programmers has been developing a new way of thinking about memory, and developing tools in the C++ language to make it easier to work in this way.

Modern C++ code rarely or never uses “delete” or “free” to deallocate memory, but instead defines clearly which object owns each other object. ...

Quick Q: How do I use std::tie and std::ignore? -- StackOverflow

By Blog Staff | May 2, 2013 01:13 PM | Tags: basics

How well do you know tie and ignore, especially to use the C++11 multiple return value idiom?

Please explain this code that uses std::ignore

I'm reading the documentation on std::ignore from cppreference. I find it quite hard to grasp the true purpose of this object, and the example code doesn't do it much justice. For example, in the below code, how and why is inserted set to true? It doesn't make much sense to me.

#include <iostream>
#include <string>
#include <set>
#include <tuple>

int main()
{
    std::set<std::string> set_of_str;
    bool inserted;
    std::tie(std::ignore, inserted) = set_of_str.insert("Test");
    if (inserted) {
        std::cout << "Value was inserted sucessfully\n";
    }
}

If someone can explain the code to me, it would be appreciated. Thanks.

Quick Q: Why does shared_ptr of void work? -- StackOverflow

By Blog Staff | May 2, 2013 11:16 AM | Tags: intermediate basics

Quick A: When it comes to destruction, only your deleter knows for sure... but he's captured at construction time, so all starts well and stays well.

Why do std::shared_ptr<void> work

I found some code using std::shared_ptr to perform arbitrary cleanup at shutdown. At first I thought this code could not possibly work, but then I tried the following:

[edited]

int main() {
  vector<shared_ptr<void>> v;
  {
    v.push_back( shared_ptr<test>( new test() ) );
  }
} // [[ how can this destroy type-safely? ]]

Quick Q: Does [=] capture all variables in scope? -- StackOverflow

By Blog Staff | May 2, 2013 04:45 AM | Tags: basics

Quick A: No.

A simple but important question:

C++11 lambda capture semantics

When I use [=] to indicate that I would like all local variables to be captured by value in a lambda, will that result in all local variables in the function being copied, or just all local variables that are used by the lambda?

So, for example, if I have:

vector<int> my_huge_vector(100000);
int my_measly_int;
some_function([=](int i){ return my_measly_int + i; });

Will my_huge_vector be copied, even though I don't use it in the lambda?

Quick Q: static constexpr variable vs. constexpr function? -- StackOverflow

By Blog Staff | May 1, 2013 04:39 PM | Tags: intermediate basics

With a nice Quick A by Morwenn that not only gives the right answer as of today, but is current with a feature voted into C++14 just two weeks ago that lets you drop the ()'s:

static constexpr variable vs. function

Is there a difference between declaring floating point constant as a static constexpr variable and a function as in example below, or is it just a matter of style?

class MY_PI
{
public:
    static constexpr float MY_PI_VAR = 3.14f;
    static constexpr float MY_PI_FUN() { return 3.14f; }
}