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    <h2>C++17 Update</h2> <p>In C++17, the meaning of <code>A_factory_func()</code> changed from creating a temporary object (C++&lt;=14) to just specifying the initialization of whatever object this expression is initialized to (loosely speaking) in C++17. These objects are the variables created by a declaration (like <code>a1</code>) or artificial objects created when the initialization ends up being discarded otherwise or if an object is needed for reference binding (like, in <code>A_factory_func();</code>, an object is artificially created, called "temporary materialization", because <code>A_factory_func()</code> doesn't have a variable or reference that otherwise would require an object to exist). Also, these objects are called "result objects".</p> <p>As examples in our case, in the case of <code>a1</code> and <code>a2</code> special rules say that in such declarations, the result object of a prvalue initializer of the same type as <code>a1</code> is variable <code>a1</code>, and therefore <code>A_factory_func()</code> directly initializes the object <code>a1</code>. Any intermediary functional-style cast would not have any effect, because <code>A_factory_func(another-prvalue)</code> just "passes through" the result object of the outer prvalue to be also the result object of the inner prvalue.</p> <hr> <pre><code>A a1 = A_factory_func(); A a2(A_factory_func()); </code></pre> <p>Depends on what type <code>A_factory_func()</code> returns. I assume it returns an <code>A</code> - then it's doing the same - except that when the copy constructor is explicit, then the first one will fail. Read <a href="http://eel.is/c++draft/dcl.init#14" rel="noreferrer">8.6/14</a></p> <pre><code>double b1 = 0.5; double b2(0.5); </code></pre> <p>This is doing the same because it's a built-in type (this means not a class type here). Read <a href="http://eel.is/c++draft/dcl.init#14" rel="noreferrer">8.6/14</a>. </p> <pre><code>A c1; A c2 = A(); A c3(A()); </code></pre> <p>This is not doing the same. The first default-initializes if <code>A</code> is a non-POD, and doesn't do any initialization for a POD (Read <a href="http://eel.is/c++draft/dcl.init#9" rel="noreferrer">8.6/9</a>). The second copy initializes: Value-initializes a temporary and then copies that value into <code>c2</code> (Read <a href="http://eel.is/c++draft/expr.type.conv#2" rel="noreferrer">5.2.3/2</a> and <a href="http://eel.is/c++draft/dcl.init#14" rel="noreferrer">8.6/14</a>). This of course will require a non-explicit copy constructor (Read <a href="http://eel.is/c++draft/dcl.init#14" rel="noreferrer">8.6/14</a> and <a href="http://eel.is/c++draft/class.conv.ctor#3" rel="noreferrer">12.3.1/3</a> and <a href="http://eel.is/c++draft/over.match.ctor#1" rel="noreferrer">13.3.1.3/1</a> ). The third creates a function declaration for a function <code>c3</code> that returns an <code>A</code> and that takes a function pointer to a function returning a <code>A</code> (Read <a href="http://eel.is/c++draft/dcl.ambig.res" rel="noreferrer">8.2</a>).</p> <hr> <p><strong>Delving into Initializations</strong> Direct and Copy initialization</p> <p>While they look identical and are supposed to do the same, these two forms are remarkably different in certain cases. The two forms of initialization are direct and copy initialization:</p> <pre><code>T t(x); T t = x; </code></pre> <p>There is behavior we can attribute to each of them:</p> <ul> <li>Direct initialization behaves like a function call to an overloaded function: The functions, in this case, are the constructors of <code>T</code> (including <code>explicit</code> ones), and the argument is <code>x</code>. Overload resolution will find the best matching constructor, and when needed will do any implicit conversion required. </li> <li>Copy initialization constructs an implicit conversion sequence: It tries to convert <code>x</code> to an object of type <code>T</code>. (It then may copy over that object into the to-initialized object, so a copy constructor is needed too - but this is not important below)</li> </ul> <p>As you see, <em>copy initialization</em> is in some way a part of direct initialization with regard to possible implicit conversions: While direct initialization has all constructors available to call, and <em>in addition</em> can do any implicit conversion it needs to match up argument types, copy initialization can just set up one implicit conversion sequence. </p> <p>I tried hard and <a href="http://coliru.stacked-crooked.com/a/708ae8b380c63ba8" rel="noreferrer">got the following code to output different text for each of those forms</a>, without using the "obvious" through <code>explicit</code> constructors.</p> <pre><code>#include &lt;iostream&gt; struct B; struct A { operator B(); }; struct B { B() { } B(A const&amp;) { std::cout &lt;&lt; "&lt;direct&gt; "; } }; A::operator B() { std::cout &lt;&lt; "&lt;copy&gt; "; return B(); } int main() { A a; B b1(a); // 1) B b2 = a; // 2) } // output: &lt;direct&gt; &lt;copy&gt; </code></pre> <p>How does it work, and why does it output that result? </p> <ol> <li><p><strong>Direct initialization</strong></p> <p>It first doesn't know anything about conversion. It will just try to call a constructor. In this case, the following constructor is available and is an <em>exact match</em>: </p> <pre><code>B(A const&amp;) </code></pre> <p>There is no conversion, much less a user defined conversion, needed to call that constructor (note that no const qualification conversion happens here either). And so direct initialization will call it. </p></li> <li><p><strong>Copy initialization</strong></p> <p>As said above, copy initialization will construct a conversion sequence when <code>a</code> has not type <code>B</code> or derived from it (which is clearly the case here). So it will look for ways to do the conversion, and will find the following candidates</p> <pre><code>B(A const&amp;) operator B(A&amp;); </code></pre> <p>Notice how I rewrote the conversion function: The parameter type reflects the type of the <code>this</code> pointer, which in a non-const member function is to non-const. Now, we call these candidates with <code>x</code> as argument. The winner is the conversion function: Because if we have two candidate functions both accepting a reference to the same type, then the <em>less const</em> version wins (this is, by the way, also the mechanism that prefers non-const member function calls for non-const objects). </p> <p>Note that if we change the conversion function to be a const member function, then the conversion is ambiguous (because both have a parameter type of <code>A const&amp;</code> then): The Comeau compiler rejects it properly, but GCC accepts it in non-pedantic mode. Switching to <code>-pedantic</code> makes it output the proper ambiguity warning too, though. </p></li> </ol> <p>I hope this helps somewhat to make it clearer how these two forms differ!</p>
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    1. POIs there a difference between copy initialization and direct initialization?
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    1. USJohannes Schaub - litb
    1. USJohannes Schaub - litb
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    1. COWow. I didn't even realize about the function declaration. I pretty much have to accept your answer just for being the only one to know about that. Is there a reason that function declarations work that way? It would be better if c3 were treated differently inside a function.
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    2. COPOD == Plain Old Data?
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    3. CO@quant_dev, yes that's it
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