Operator [] overloading - c++

I have the following code:
class A
{
public:
A() {};
void operator[](int x)
{
}
};
int _tmain(int argc, _TCHAR* argv[])
{
A a;
a.operator[](0);
a[0];
}
Both calls work, but I want to know whether there is any difference. Is one more efficient than the other? Do other things happen(besides executing the code in the overloaded operator) in either case?
EDIT:
Is there a case why you would want to write a.operator instead of just []. What's the point of overloading if you're not using the short syntax?

Both calls are identical. All the operators can be called with an explicit .operator## syntax, where ## stands for the actual operator symbol.
It is the literal operators like a + b that are just syntactic sugar for a.operator+(b) when it comes to classes. (Though of course for primitive types that is not the case.)
Note that your []-operator is not very typical, since it is usually used to return a reference to something -- but it's up to you how to implement it. You're restricted to one single argument, though, unlike operator().

The explicit operator[] (and any other explicit operator) is used in an inheritence heirarchy, where you are trying to call operator[] on a base class. You can't do that using the normal [], as it would result in a recursive function call. ie; you might use something like this:
struct Base {
void operator[] (int i) { }
};
struct Derived : public Base {
void operator[] (int i)
{ Base::operator[](i); }
};

Other than having to type another 11 characters, there is no functional difference between the two.

They are completely equivalent - just in once case C++ adds syntactic sugar that makes your code prettier. You can call all overloaded operators explicitly - the effect will be the same as when they are called implicitly and the compiler redirects calls to them.

No there is no difference between both versions from performance point of view. a[0] is more readable.
Also, the operator [] is typically defined as,
Type& operator[](const int x)
{
return v[x]; // Type 'v' is member variable which you are accessing for array
}

Related

Overloading parentheses operator on vector in c++

I've tried reading several of the overloading questions here to get an idea how to do that, but as I understand overloading parentheses is different from other operators as it needs to be overloaded inside the class?
I have one file in my project, main.cpp, and I'm trying to overload the () operator as follows:
class vector<int> {
public:
bool operator((iterator a);
};
With a matching function.
bool vector<int>::operator()(vector<int>::iterator a) {
return (*a > 0);
}
But I get several errors, the first one being:
an explicit specialization must be preceded by 'template <>' class
vector {
I've tried to correct what the errors ask for, but it seems my understanding of the process is just not good enough.
What would be the correct method of overloading the operator here?
Thanks in advance for any replies.
You have read right: operator() is one of the four operators (along with =, [] and ->) that can only be implemented as class members. And since std::vector is not yours (be it te template itself or any class specialized from it), you cannot implement them for it.
There is still a solution though, and that is to wrap std::vector inside a class of your own, and overload operator() for that:
struct callableIntVector : std::vector<int> {
using std::vector<int>::vector;
bool operator ()(std::vector<int>::iterator a) const {
return *a > 0;
}
};
The usual caveats about inheriting from standard containers apply: don't destruct them polymorphically as they have no virtual destructor, take care not to slice them, etc.

When an object provides both `operator!` and `operator bool`, which is used in the expression `!obj`?

I've ran across a question I am not able to answer for myself. Also, I didn't find an answer to this on both google and here. Say, I want to "check an object for validity" in an if clause, like so:
MyClass myObject;
// [some code, if any]
if (!myObject)
{
// [do something]
}
Let MyClass be defined something like this:
class MyClass
{
public:
MyClass() { };
virtual ~MyClass() { };
bool operator!()
{
return !myBool;
};
operator bool()
{
return myBool;
};
private:
bool myBool = 0;
};
My question now is: Which one of the overloaded operators is actually used in this if clause? Either way, the result is obviously the same.
It will use operator!.
A function whose parameter types match the arguments will be chosen in preference to one that requires type conversions.
You'll find that operator ! gets executed because it's the most direct resolution. If it used operator bool instead then it would have to call the conversion operator first, and then apply the ! separately to that.
As a general rule, it's a good idea to avoid that situation though. It's generally better just to define the bool conversion, because strictly speaking that's what you want your logical operators to act on, rather than MyClass directly. Defining both creates a bit of a readability problem, and is a form of redundant code duplication (which can lead to programmer error in future).

The Arrow Member Operator in C++

I am quite new to using C++. I have handled Java and ActionScript before, but now I want to learn this powerful language. Since C++ grants the programmer the ability to explicitly use pointers, I am quite confused over the use of the arrow member operator. Here is a sample code I tried writing.
main.cpp:
#include <iostream>
#include "Arrow.h"
using namespace std;
int main()
{
Arrow object;
Arrow *pter = &object;
object.printCrap(); //Using Dot Access
pter->printCrap(); //Using Arrow Member Operator
return 0;
}
Arrow.cpp
#include <iostream>
#include "Arrow.h"
using namespace std;
Arrow::Arrow()
{
}
void Arrow::printCrap(){
cout << "Steak!" << endl;
}
In the above code, all it does is to print steak, using both methods (Dot and Arrow).
In short, in writing a real practical application using C++, when do I use the arrow notation? I’m used to using the dot notation due to my previous programming experience, but the arrow is completely new to me.
In C, a->b is precisely equivalent to (*a).b. The "arrow" notation was introduced as a convenience; accessing a member of a struct via a pointer was fairly common and the arrow notation is easier to write/type, and generally considered more readable as well.
C++ adds another wrinkle as well though: operator-> can be overloaded for a struct/class. Although fairly unusual otherwise, doing so is common (nearly required) for smart pointer classes.
That's not really unusual in itself: C++ allows the vast majority of operators to be overloaded (although some almost never should be, such as operator&&, operator|| and operator,).
What is unusual is how an overloaded operator-> is interpreted. First, although a->b looks like -> is a binary operator, when you overload it in C++, it's treated as a unary operator, so the correct signature is T::operator(), not T::operator(U) or something on that order.
The result is interpreted somewhat unusually as well. Assuming foo is an object of some type that overloads operator->, foo->bar is interpreted as meaning (f.operator->())->bar. That, in turn, restricts the return type of an overloaded operator->. Specifically, it must return either an instance of another class that also overloads operator-> (or a reference to such an object) or else it must return a pointer.
In the former case, a simple-looking foo->bar could actually mean "chasing" through an entire (arbitrarily long) chain of instances of objects, each of which overloads operator->, until one is finally reached that can refer to a member named bar. For an (admittedly extreme) example, consider this:
#include <iostream>
class int_proxy {
int val;
public:
int_proxy(): val(0) {}
int_proxy &operator=(int n) {
std::cout<<"int_proxy::operator= called\n";
val=n;
return *this;
}
};
struct fubar {
int_proxy bar;
} instance;
struct h {
fubar *operator->() {
std::cout<<"used h::operator->\n";
return &instance;
}
};
struct g {
h operator->() {
std::cout<<"used g::operator->\n";
return h();
}
};
struct f {
g operator->() {
std::cout<<"Used f::operator->\n";
return g();
}
};
int main() {
f foo;
foo->bar=1;
}
Even though foo->bar=1; looks like a simple assignment to a member via a pointer, this program actually produces the following output:
Used f::operator->
used g::operator->
used h::operator->
int_proxy::operator= called
Clearly, in this case foo->bar is not (even close to) equivalent to a simple (*foo).bar. As is obvious from the output, the compiler generates "hidden" code to walk through the whole series of overloaded -> operators in various classes to get from foo to (a pointer to) something that has a member named bar (which in this case is also a type that overloads operator=, so we can see output from the assignment as well).
Good Question,
Dot(.) this operator is used for accessing the member function or
sometime the data member of a class or structure using instance
variable of that class/Structure.
object.function();
object.dataMember; //not a standard for class.
arrow(->) this operator is used for accessing the member function or
sometime the data member of a class or structure but using pointer of
that class/Structure.
ptr->function();
ptr->datamember; //not a standard for class.
The -> operator is a way of calling a member function of the pointer that is being dereferenced. It can also be written as (*pter).printCap(). C++ is difficult to learn without a class or book so I recommend getting one, it'll be a great investement!

Operator overloading in template in C++

I read following code from somewhere:
template<class T> class A {
T a;
public:
A(T x):a(x) {}
operator T() const {return a;} // what is point here?
};
int _tmain(int argc, _TCHAR* argv[])
{
A<int> a = A<int>(5);
int n = a;
cout << n;
return 0;
}
What does below line mean?
operator T() const {return a;}
operator T() const {return a;}
This is the typecast operator. It'll implicitly convert the class instance to T. In the example code you've posted this conversion is being performed at the line
int n = a;
It means if you want to convert an instance into a T you can use this operator, which here returns a copy of the private member.
In your example code that is how you can assign a, which is of type A<int> to an int directly. Try removing the operator T() and see how that fails to compile, with an error about assigining A<T> to an int.
With the non explicit constructor (the opposite of marking a constructor explicit) there too it makes this type behave a lot like the template type itself in a number of circumstances. In effect you've wrapped a T inside another class that behaves like a T when it needs to. You could extend this to do other, more useful things like monitoring/logging/restricting the use of real instances by hiding them behind something which controlled them.
Also notice how you can change A<int> a = A<int>(5); to simply A<int> a = 5; because of the implicit constructor.
It's basically making a functor - which is an object with function semantics. That means you can call the object just like a function as a replacement in places where you may have used a function - its a generic programming concept.
It's beneficial because you can have multiple instances of that function-object (functor) and they can each maintain their own state, where as if you had a straight-up function then it could only maintain state via static variables, and it would thus not be re-entrant (you only ever get one instance of a static variable).
Functors are heavily used in STL algorithms as an extra optional parameter.

Hidden Features of C++? [closed]

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No C++ love when it comes to the "hidden features of" line of questions? Figured I would throw it out there. What are some of the hidden features of C++?
Most C++ programmers are familiar with the ternary operator:
x = (y < 0) ? 10 : 20;
However, they don't realize that it can be used as an lvalue:
(a == 0 ? a : b) = 1;
which is shorthand for
if (a == 0)
a = 1;
else
b = 1;
Use with caution :-)
You can put URIs into C++ source without error. For example:
void foo() {
http://stackoverflow.com/
int bar = 4;
...
}
Pointer arithmetics.
C++ programmers prefer to avoid pointers because of the bugs that can be introduced.
The coolest C++ I've ever seen though? Analog literals.
I agree with most posts there: C++ is a multi-paradigm language, so the "hidden" features you'll find (other than "undefined behaviours" that you should avoid at all cost) are clever uses of facilities.
Most of those facilities are not build-in features of the language, but library-based ones.
The most important is the RAII, often ignored for years by C++ developers coming from the C world. Operator overloading is often a misunderstood feature that enable both array-like behaviour (subscript operator), pointer like operations (smart pointers) and build-in-like operations (multiplying matrices.
The use of exception is often difficult, but with some work, can produce really robust code through exception safety specifications (including code that won't fail, or that will have a commit-like features that is that will succeed, or revert back to its original state).
The most famous of "hidden" feature of C++ is template metaprogramming, as it enables you to have your program partially (or totally) executed at compile-time instead of runtime. This is difficult, though, and you must have a solid grasp on templates before trying it.
Other make uses of the multiple paradigm to produce "ways of programming" outside of C++'s ancestor, that is, C.
By using functors, you can simulate functions, with the additional type-safety and being stateful. Using the command pattern, you can delay code execution. Most other design patterns can be easily and efficiently implemented in C++ to produce alternative coding styles not supposed to be inside the list of "official C++ paradigms".
By using templates, you can produce code that will work on most types, including not the one you thought at first. You can increase type safety,too (like an automated typesafe malloc/realloc/free). C++ object features are really powerful (and thus, dangerous if used carelessly), but even the dynamic polymorphism have its static version in C++: the CRTP.
I have found that most "Effective C++"-type books from Scott Meyers or "Exceptional C++"-type books from Herb Sutter to be both easy to read, and quite treasures of info on known and less known features of C++.
Among my preferred is one that should make the hair of any Java programmer rise from horror: In C++, the most object-oriented way to add a feature to an object is through a non-member non-friend function, instead of a member-function (i.e. class method), because:
In C++, a class' interface is both its member-functions and the non-member functions in the same namespace
non-friend non-member functions have no privileged access to the class internal. As such, using a member function over a non-member non-friend one will weaken the class' encapsulation.
This never fails to surprise even experienced developers.
(Source: Among others, Herb Sutter's online Guru of the Week #84: http://www.gotw.ca/gotw/084.htm )
One language feature that I consider to be somewhat hidden, because I had never heard about it throughout my entire time in school, is the namespace alias. It wasn't brought to my attention until I ran into examples of it in the boost documentation. Of course, now that I know about it you can find it in any standard C++ reference.
namespace fs = boost::filesystem;
fs::path myPath( strPath, fs::native );
Not only can variables be declared in the init part of a for loop, but also classes and functions.
for(struct { int a; float b; } loop = { 1, 2 }; ...; ...) {
...
}
That allows for multiple variables of differing types.
The array operator is associative.
A[8] is a synonym for *(A + 8). Since addition is associative, that can be rewritten as *(8 + A), which is a synonym for..... 8[A]
You didn't say useful... :-)
One thing that's little known is that unions can be templates too:
template<typename From, typename To>
union union_cast {
From from;
To to;
union_cast(From from)
:from(from) { }
To getTo() const { return to; }
};
And they can have constructors and member functions too. Just nothing that has to do with inheritance (including virtual functions).
C++ is a standard, there shouldn't be any hidden features...
C++ is a multi-paradigm language, you can bet your last money on there being hidden features. One example out of many: template metaprogramming. Nobody in the standards committee intended there to be a Turing-complete sublanguage that gets executed at compile-time.
Another hidden feature that doesn't work in C is the functionality of the unary + operator. You can use it to promote and decay all sorts of things
Converting an Enumeration to an integer
+AnEnumeratorValue
And your enumerator value that previously had its enumeration type now has the perfect integer type that can fit its value. Manually, you would hardly know that type! This is needed for example when you want to implement an overloaded operator for your enumeration.
Get the value out of a variable
You have to use a class that uses an in-class static initializer without an out of class definition, but sometimes it fails to link? The operator may help to create a temporary without making assumptins or dependencies on its type
struct Foo {
static int const value = 42;
};
// This does something interesting...
template<typename T>
void f(T const&);
int main() {
// fails to link - tries to get the address of "Foo::value"!
f(Foo::value);
// works - pass a temporary value
f(+Foo::value);
}
Decay an array to a pointer
Do you want to pass two pointers to a function, but it just won't work? The operator may help
// This does something interesting...
template<typename T>
void f(T const& a, T const& b);
int main() {
int a[2];
int b[3];
f(a, b); // won't work! different values for "T"!
f(+a, +b); // works! T is "int*" both time
}
Lifetime of temporaries bound to const references is one that few people know about. Or at least it's my favorite piece of C++ knowledge that most people don't know about.
const MyClass& x = MyClass(); // temporary exists as long as x is in scope
A nice feature that isn't used often is the function-wide try-catch block:
int Function()
try
{
// do something here
return 42;
}
catch(...)
{
return -1;
}
Main usage would be to translate exception to other exception class and rethrow, or to translate between exceptions and return-based error code handling.
Many know of the identity / id metafunction, but there is a nice usecase for it for non-template cases: Ease writing declarations:
// void (*f)(); // same
id<void()>::type *f;
// void (*f(void(*p)()))(int); // same
id<void(int)>::type *f(id<void()>::type *p);
// int (*p)[2] = new int[10][2]; // same
id<int[2]>::type *p = new int[10][2];
// void (C::*p)(int) = 0; // same
id<void(int)>::type C::*p = 0;
It helps decrypting C++ declarations greatly!
// boost::identity is pretty much the same
template<typename T>
struct id { typedef T type; };
A quite hidden feature is that you can define variables within an if condition, and its scope will span only over the if, and its else blocks:
if(int * p = getPointer()) {
// do something
}
Some macros use that, for example to provide some "locked" scope like this:
struct MutexLocker {
MutexLocker(Mutex&);
~MutexLocker();
operator bool() const { return false; }
private:
Mutex &m;
};
#define locked(mutex) if(MutexLocker const& lock = MutexLocker(mutex)) {} else
void someCriticalPath() {
locked(myLocker) { /* ... */ }
}
Also BOOST_FOREACH uses it under the hood. To complete this, it's not only possible in an if, but also in a switch:
switch(int value = getIt()) {
// ...
}
and in a while loop:
while(SomeThing t = getSomeThing()) {
// ...
}
(and also in a for condition). But i'm not too sure whether these are all that useful :)
Preventing comma operator from calling operator overloads
Sometimes you make valid use of the comma operator, but you want to ensure that no user defined comma operator gets into the way, because for instance you rely on sequence points between the left and right side or want to make sure nothing interferes with the desired action. This is where void() comes into game:
for(T i, j; can_continue(i, j); ++i, void(), ++j)
do_code(i, j);
Ignore the place holders i put for the condition and code. What's important is the void(), which makes the compiler force to use the builtin comma operator. This can be useful when implementing traits classes, sometimes, too.
Array initialization in constructor.
For example in a class if we have a array of int as:
class clName
{
clName();
int a[10];
};
We can initialize all elements in the array to its default (here all elements of array to zero) in the constructor as:
clName::clName() : a()
{
}
Oooh, I can come up with a list of pet hates instead:
Destructors need to be virtual if you intend use polymorphically
Sometimes members are initialized by default, sometimes they aren't
Local clases can't be used as template parameters (makes them less useful)
exception specifiers: look useful, but aren't
function overloads hide base class functions with different signatures.
no useful standardisation on internationalisation (portable standard wide charset, anyone? We'll have to wait until C++0x)
On the plus side
hidden feature: function try blocks. Unfortunately I haven't found a use for it. Yes I know why they added it, but you have to rethrow in a constructor which makes it pointless.
It's worth looking carefully at the STL guarantees about iterator validity after container modification, which can let you make some slightly nicer loops.
Boost - it's hardly a secret but it's worth using.
Return value optimisation (not obvious, but it's specifically allowed by the standard)
Functors aka function objects aka operator(). This is used extensively by the STL. not really a secret, but is a nifty side effect of operator overloading and templates.
You can access protected data and function members of any class, without undefined behavior, and with expected semantics. Read on to see how. Read also the defect report about this.
Normally, C++ forbids you to access non-static protected members of a class's object, even if that class is your base class
struct A {
protected:
int a;
};
struct B : A {
// error: can't access protected member
static int get(A &x) { return x.a; }
};
struct C : A { };
That's forbidden: You and the compiler don't know what the reference actually points at. It could be a C object, in which case class B has no business and clue about its data. Such access is only granted if x is a reference to a derived class or one derived from it. And it could allow arbitrary piece of code to read any protected member by just making up a "throw-away" class that reads out members, for example of std::stack:
void f(std::stack<int> &s) {
// now, let's decide to mess with that stack!
struct pillager : std::stack<int> {
static std::deque<int> &get(std::stack<int> &s) {
// error: stack<int>::c is protected
return s.c;
}
};
// haha, now let's inspect the stack's middle elements!
std::deque<int> &d = pillager::get(s);
}
Surely, as you see this would cause way too much damage. But now, member pointers allow circumventing this protection! The key point is that the type of a member pointer is bound to the class that actually contains said member - not to the class that you specified when taking the address. This allows us to circumvent checking
struct A {
protected:
int a;
};
struct B : A {
// valid: *can* access protected member
static int get(A &x) { return x.*(&B::a); }
};
struct C : A { };
And of course, it also works with the std::stack example.
void f(std::stack<int> &s) {
// now, let's decide to mess with that stack!
struct pillager : std::stack<int> {
static std::deque<int> &get(std::stack<int> &s) {
return s.*(pillager::c);
}
};
// haha, now let's inspect the stack's middle elements!
std::deque<int> &d = pillager::get(s);
}
That's going to be even easier with a using declaration in the derived class, which makes the member name public and refers to the member of the base class.
void f(std::stack<int> &s) {
// now, let's decide to mess with that stack!
struct pillager : std::stack<int> {
using std::stack<int>::c;
};
// haha, now let's inspect the stack's middle elements!
std::deque<int> &d = s.*(&pillager::c);
}
Another hidden feature is that you can call class objects that can be converted to function pointers or references. Overload resolution is done on the result of them, and arguments are perfectly forwarded.
template<typename Func1, typename Func2>
class callable {
Func1 *m_f1;
Func2 *m_f2;
public:
callable(Func1 *f1, Func2 *f2):m_f1(f1), m_f2(f2) { }
operator Func1*() { return m_f1; }
operator Func2*() { return m_f2; }
};
void foo(int i) { std::cout << "foo: " << i << std::endl; }
void bar(long il) { std::cout << "bar: " << il << std::endl; }
int main() {
callable<void(int), void(long)> c(foo, bar);
c(42); // calls foo
c(42L); // calls bar
}
These are called "surrogate call functions".
Hidden features:
Pure virtual functions can have implementation. Common example, pure virtual destructor.
If a function throws an exception not listed in its exception specifications, but the function has std::bad_exception in its exception specification, the exception is converted into std::bad_exception and thrown automatically. That way you will at least know that a bad_exception was thrown. Read more here.
function try blocks
The template keyword in disambiguating typedefs in a class template. If the name of a member template specialization appears after a ., ->, or :: operator, and that name has explicitly qualified template parameters, prefix the member template name with the keyword template. Read more here.
function parameter defaults can be changed at runtime. Read more here.
A[i] works as good as i[A]
Temporary instances of a class can be modified! A non-const member function can be invoked on a temporary object. For example:
struct Bar {
void modify() {}
}
int main (void) {
Bar().modify(); /* non-const function invoked on a temporary. */
}
Read more here.
If two different types are present before and after the : in the ternary (?:) operator expression, then the resulting type of the expression is the one that is the most general of the two. For example:
void foo (int) {}
void foo (double) {}
struct X {
X (double d = 0.0) {}
};
void foo (X) {}
int main(void) {
int i = 1;
foo(i ? 0 : 0.0); // calls foo(double)
X x;
foo(i ? 0.0 : x); // calls foo(X)
}
map::operator[] creates entry if key is missing and returns reference to default-constructed entry value. So you can write:
map<int, string> m;
string& s = m[42]; // no need for map::find()
if (s.empty()) { // assuming we never store empty values in m
s.assign(...);
}
cout << s;
I'm amazed at how many C++ programmers don't know this.
Putting functions or variables in a nameless namespace deprecates the use of static to restrict them to file scope.
Defining ordinary friend functions in class templates needs special attention:
template <typename T>
class Creator {
friend void appear() { // a new function ::appear(), but it doesn't
… // exist until Creator is instantiated
}
};
Creator<void> miracle; // ::appear() is created at this point
Creator<double> oops; // ERROR: ::appear() is created a second time!
In this example, two different instantiations create two identical definitions—a direct violation of the ODR
We must therefore make sure the template parameters of the class template appear in the type of any friend function defined in that template (unless we want to prevent more than one instantiation of a class template in a particular file, but this is rather unlikely). Let's apply this to a variation of our previous example:
template <typename T>
class Creator {
friend void feed(Creator<T>*){ // every T generates a different
… // function ::feed()
}
};
Creator<void> one; // generates ::feed(Creator<void>*)
Creator<double> two; // generates ::feed(Creator<double>*)
Disclaimer: I have pasted this section from C++ Templates: The Complete Guide / Section 8.4
void functions can return void values
Little known, but the following code is fine
void f() { }
void g() { return f(); }
Aswell as the following weird looking one
void f() { return (void)"i'm discarded"; }
Knowing about this, you can take advantage in some areas. One example: void functions can't return a value but you can also not just return nothing, because they may be instantiated with non-void. Instead of storing the value into a local variable, which will cause an error for void, just return a value directly
template<typename T>
struct sample {
// assume f<T> may return void
T dosomething() { return f<T>(); }
// better than T t = f<T>(); /* ... */ return t; !
};
Read a file into a vector of strings:
vector<string> V;
copy(istream_iterator<string>(cin), istream_iterator<string>(),
back_inserter(V));
istream_iterator
You can template bitfields.
template <size_t X, size_t Y>
struct bitfield
{
char left : X;
char right : Y;
};
I have yet to come up with any purpose for this, but it sure as heck surprised me.
One of the most interesting grammars of any programming languages.
Three of these things belong together, and two are something altogether different...
SomeType t = u;
SomeType t(u);
SomeType t();
SomeType t;
SomeType t(SomeType(u));
All but the third and fifth define a SomeType object on the stack and initialize it (with u in the first two case, and the default constructor in the fourth. The third is declaring a function that takes no parameters and returns a SomeType. The fifth is similarly declaring a function that takes one parameter by value of type SomeType named u.
Getting rid of forward declarations:
struct global
{
void main()
{
a = 1;
b();
}
int a;
void b(){}
}
singleton;
Writing switch-statements with ?: operators:
string result =
a==0 ? "zero" :
a==1 ? "one" :
a==2 ? "two" :
0;
Doing everything on a single line:
void a();
int b();
float c = (a(),b(),1.0f);
Zeroing structs without memset:
FStruct s = {0};
Normalizing/wrapping angle- and time-values:
int angle = (short)((+180+30)*65536/360) * 360/65536; //==-150
Assigning references:
struct ref
{
int& r;
ref(int& r):r(r){}
};
int b;
ref a(b);
int c;
*(int**)&a = &c;
The ternary conditional operator ?: requires its second and third operand to have "agreeable" types (speaking informally). But this requirement has one exception (pun intended): either the second or third operand can be a throw expression (which has type void), regardless of the type of the other operand.
In other words, one can write the following pefrectly valid C++ expressions using the ?: operator
i = a > b ? a : throw something();
BTW, the fact that throw expression is actually an expression (of type void) and not a statement is another little-known feature of C++ language. This means, among other things, that the following code is perfectly valid
void foo()
{
return throw something();
}
although there's not much point in doing it this way (maybe in some generic template code this might come handy).
The dominance rule is useful, but little known. It says that even if in a non-unique path through a base-class lattice, name-lookup for a partially hidden member is unique if the member belongs to a virtual base-class:
struct A { void f() { } };
struct B : virtual A { void f() { cout << "B!"; } };
struct C : virtual A { };
// name-lookup sees B::f and A::f, but B::f dominates over A::f !
struct D : B, C { void g() { f(); } };
I've used this to implement alignment-support that automatically figures out the strictest alignment by means of the dominance rule.
This does not only apply to virtual functions, but also to typedef names, static/non-virtual members and anything else. I've seen it used to implement overwritable traits in meta-programs.