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Diagnosable validity -- Andrzej KrzemieĊ„ski

By Felix Petriconi | May 17, 2016 05:24 AM | Tags: c++14 c++11 advanced

Andrzej Krzemieński wrote down his thoughts on ill-formed C++ code.


Diagnosable validity

by Andrzej Krzemieński

From the article:

Certain combinations of types and expressions can make a C++ program ill-formed. “Ill-formed” is a term taken from the C++ Standard and it means that a program is not valid, and compiler must (in most of the cases) reject it. This is quite obvious:

int main()
{
  auto i = "some text".size(); // invalid expression
};

String literals do not have member functions, therefore compiler cannot accept this program, and must report an error. This puts a responsibility on programmers to learn which expressions and types are valid in a given context and use only these. Again, I am saying a very obvious thing.

What is less obvious is that there is a way in C++ to enter a type or expression of which we do not know if it is valid or not, in an isolated environment, where it does not render the entire program ill-formed, but instead it returns a yes-no (or rather valid-invalid) answer, which we can use at compile-time to make a decision how we want the program to behave. When requested, compiler can analyze all the declarations it has seen so far, and make an approximated judgement whether a given type or expression would make the program ill-formed or not, if used outside the isolated environment. The compiler’s approximated answer is not always correct, but it is just enough most of the time.

 

Quick Q: Templated Function results in Circular Inclusion

By Adrien Hamelin | May 16, 2016 01:36 PM | Tags: basics

Quick A: Separate definition and implementation.

Recently on SO:

Templated Function results in Circular Inclusion

Define registerEvent after IApp.

class IApp;

class Component
{
    IApp* app;
    template<typename T>
    void registerEvent(const int& evtId, Status (T::*func) (int));
};

class IApp : public Component {
  ...
};

template <typename T>
Component::registerEvent(const int& evtId, Status (T::*func) (int)) {
  auto res = std::bind(func, (T*)this, std::placeholders::_1);
  app->registerForEvent(evtId);
}

If need be, also define A::registerEvent after Component::registerEvent.

Quick Q: Why is list initialization (using curly braces) better than the alternatives?

By Adrien Hamelin | May 16, 2016 01:31 PM | Tags: c++11 basics

Quick A: It is less likely to generate an unexpected error.

Recently on SO:

Why is list initialization (using curly braces) better than the alternatives?

Basically copying and pasting from Bjarne Stroustrup's "The C++ Programming Language 4th Edition":

List initialization does not allow narrowing (§iso.8.5.4). That is:

  • An integer cannot be converted to another integer that cannot hold its value. For example, char to int is allowed, but not int to char.
  • A floating-point value cannot be converted to another floating-point type that cannot hold its value. For example, float to double is allowed, but not double to float.
  • A floating-point value cannot be converted to an integer type.
  • An integer value cannot be converted to a floating-point type.

Example:

void fun(double val, int val2) {

    int x2 = val; // if val==7.9, x2 becomes 7 (bad)

    char c2 = val2; // if val2==1025, c2 becomes 1 (bad)

    int x3 {val}; // error: possible truncation (good)

    char c3 {val2}; // error: possible narrowing (good)

    char c4 {24}; // OK: 24 can be represented exactly as a char (good)

    char c5 {264}; // error (assuming 8-bit chars): 264 cannot be
                   // represented as a char (good)

    int x4 {2.0}; // error: no double to int value conversion (good)

}

The only situation where = is preferred over {} is when using auto keyword to get the type determined by the initializer.

Example:

auto z1 {99}; // z1 is an initializer_list<int>
auto z2 = 99; // z2 is an int

Conclusion

Prefer {} initialization over alternatives unless you have a strong reason not to.

Top 10 dumb mistakes to avoid with C++ 11 smart pointers -- Deb Haldar

By Mantosh Kumar | May 15, 2016 10:36 PM | Tags: intermediate

Discussion on various aspect of C++11 smart pointers uses.

Top 10 dumb mistakes to avoid with C++ 11 smart pointers

by  Deb Haldar

From the article:

I love the new C++ 11 smart pointers. In many ways, they were a godsent for many folks who hate managing their own memory. In my opinion, it made teaching C++ to newcomers much easier.

However, in the two plus years that I've been using them extensively, I've come across multiple cases where improper use of the C++ 11 smart pointers made the program inefficient or simply crash and burn. I've catalogued them below for easy reference.

Quick Q: Is this std::ref behaviour logical?

By Adrien Hamelin | May 9, 2016 01:19 PM | Tags: intermediate c++11

Quick A: Yes, std::ref can be reassigned (see example)

Recently on SO:

Is this std::ref behaviour logical?

A small modification to f2 provides the clue:

template<class T>
void f2(T arg)
{
    arg.get() = xx;
}

This now does what you expect.

This has happened because std::ref returns a std::reference_wrapper<> object. The assignment operator of which rebinds the wrapper. (see http://en.cppreference.com/w/cpp/utility/functional/reference_wrapper/operator%3D)

It does not make an assignment to the wrapped reference.

In the f1 case, all is working as you expected because a std::reference_wrapper<T> provides a conversion operator to T&, which will bind to the implicit right hand side of ints implicit operator+.

Strict Weak Ordering and STL -- Saurabh Singh

By bashrc | May 2, 2016 08:17 AM | Tags: basics

Saurabh Singh describes in a brief tutorial on how to correctly implement the comparator function for STL containers and algorithms. 

Strict Weak Ordering and STL

by Saurabh Singh

From the article:

If you had ever used a map or set or even std::sort I bet you would have to give a comparator function. (Or an overload to the < (less than) operator).
I will try to give an overview of how certain associative stl containers use this property for ordering the elements.
Almost all stl containers rely on strict weak ordering A strict weak ordering defines the relative position of elements in terms of precedence of one item over other. For eg. if you have a room full of person and you have to form a queue based on their height, a person with "lesser" height will "precede" the person with greater height. For a function to be satisfying strict weak ordering following conditions need to be met.

The PDF version of the document is available here.

Quick Q: What are rvalues, lvalues, xvalues, glvalues, and prvalues?

By Adrien Hamelin | Apr 28, 2016 12:57 PM | Tags: intermediate c++11

Quick A: Categories of values that determine what can be done with that value.

Some time ago on SO:

What are rvalues, lvalues, xvalues, glvalues, and prvalues?

What are these new categories of expressions?
The FCD (n3092) has an excellent description:

— An lvalue (so called, historically, because lvalues could appear on the left-hand side of an assignment expression) designates a function or an object. [ Example: If E is an expression of pointer type, then *E is an lvalue expression referring to the object or function to which E points. As another example, the result of calling a function whose return type is an lvalue reference is an lvalue. —end example ]

— An xvalue (an “eXpiring” value) also refers to an object, usually near the end of its lifetime (so that its resources may be moved, for example). An xvalue is the result of certain kinds of expressions involving rvalue references (8.3.2). [ Example: The result of calling a function whose return type is an rvalue reference is an xvalue. —end example ]

— A glvalue (“generalized” lvalue) is an lvalue or an xvalue.

— An rvalue (so called, historically, because rvalues could appear on the right-hand side of an assignment expressions) is an xvalue, a temporary object (12.2) or subobject thereof, or a value that is not associated with an object.

— A prvalue (“pure” rvalue) is an rvalue that is not an xvalue. [ Example: The result of calling a function whose return type is not a reference is a prvalue. The value of a literal such as 12, 7.3e5, or true is also a prvalue. —end example ]

Every expression belongs to exactly one of the fundamental classifications in this taxonomy: lvalue, xvalue, or prvalue. This property of an expression is called its value category. [ Note: The discussion of each built-in operator in Clause 5 indicates the category of the value it yields and the value categories of the operands it expects. For example, the built-in assignment operators expect that the left operand is an lvalue and that the right operand is a prvalue and yield an lvalue as the result. User-defined operators are functions, and the categories of values they expect and yield are determined by their parameter and return types. —end note

I suggest you read the entire section 3.10 Lvalues and rvalues though.
How do these new categories relate to the existing rvalue and lvalue categories?
Again:

Are the rvalue and lvalue categories in C++0x the same as they are in C++03?
The semantics of rvalues has evolved particularly with the introduction of move semantics.
Why are these new categories needed?
So that move construction/assignment could be defined and supported.

Quick Q: Does C++ final imply final in all aspects?

By Adrien Hamelin | Apr 28, 2016 12:47 PM | Tags: intermediate c++11

Quick A: Yes, a final class cannot have its methods overriden.

Recently on SO:

Does C++ final imply final in all aspects?

To quote the draft C++ standard from here [class.virtual/4]:

If a virtual function f in some class B is marked with the virt-specifier final and in a class D derived from B a function D::f overrides B::f, the program is ill-formed.
And here [class/3]:
If a class is marked with the class-virt-specifier final and it appears as a base-type-specifier in a base-clause (Clause [class.derived]), the program is ill-formed.
So, in answer to the question;

 

Does a final class implicitly imply its virtual functions to be final as well? Should it? Please clarify.

So, at least not formally. But attempts to violate either rule will be the same result in both cases; the program is ill-formed and so won't compile. A final class means the class cannot be derived from, so as a consequence of this, its virtual methods cannot be overridden.

 

Should it? Probably not, they are related but they not the same thing. There is also no need formally require the one to imply the other, the effect follows naturally. Any violations have the same result, a failed compile (hopefully with appropriate error messages to distinguish the two).