A question dawned on me lately: Why bother having the scope resolution operator (::) when the usage of the dot operator (.) doesn't clash with the former?
Like,
namespace my_module
{
const int insane_constant = 69;
int some_threshold = 100;
}
In order to access some_threshold, one would write my_module::some_threshold. But the thing is the left hand side of :: must be a namespace or a class, as far as I know, while that of the . can never be a namespace or a class. Moreover, the order of :: is higher than ., which is really pointless in my opinion. Why make 2 distinct operators, both of which cannot be overloaded, with use cases can be covered with just one? I'm having a hard time finding any satisfying answer, so any clue on this matter is much appriciated.
Consider the following code:
#include <iostream>
namespace my_module
{
const int insane_constant = 69;
}
class C{
public:
int insane_constant; // actually not constant, but for the example's sake
};
int main(){
C my_module;
my_module.insane_constant = my_module::insane_constant;
std::cout << my_module.insane_constant;
}
How would you distinguish my_module. from my_module:: here?
You can't replace one with the other.
Example:
namespace A
{
int x = 0;
struct X { int x = 1; } A;
void f()
{
std::cout << A.x
<< A::x
<< std::endl;
}
}
int main()
{
A::f();
}
prints "10".
Both operators have completely different purposes.
In particular, :: is for referring to a member of a scope while operator. is for referring to a member of an object.
That is, operator:: works on scope while operator. works on objects.
For instance, a classname::membername refers to the member of a class which has a class scope associated with it. On the other hand, object.membername refers to the member of the object object.
Note that not all scopes can be used with operator::. For example, we can refer to a namespace scope's member, class scopes' member(like a static member etc), enumeration' member etc using operator::.
The other answers showcase situations where :: and . do different things because they act on different entities (scopes vs objects). I'd like to add a bit of motivation behind this choice in the language by comparing it with the situation in C#. Disclaimer: I don't actually use C#, but these points seem obvious when looking into it.
In C#, the "namespace operator" is a dot ., just like the member access operator. However, you cannot have truly global variables: you need a class to put them in. In fact, you cannot even declare a class without a namespace, though this is unrelated to your question.
In C#, there is no possible ambiguity when using the dot between scope and object. Nested namespaces, classes within namespaces, class static variables, enumerators, and so on, are all accessed with a dot.
In C++, that's obviously not the case, and the language took the route of distinguishing between scopes and objects by having two distinct operators.
In plain C, there is no scope resolution, so there is no need for a :: operator. Other languages that make the distinction are Ruby and, apparently, PHP. Among languages that do not make the distinction, you can find for example Python, Go and Java. That last one does have a :: operator, but it's not for scope resolution, it's for forming method references.
In short, C++ has complex scope mechanisms because that's how the language was designed. As a result, it requires two different operators to distinguish scope access and object access.
Why is operator () of stateless functor not allowed to be static? Stateless lambda objects are convertible to pointers to free functions having the same signature as their operator ().
Stephan T. Lavavej on p. 6 points out that conversion to a function pointer is just an operator FunctionPointer() (cite). But I can't obtain a corresponding pointer to operator () as to non-member function. For functor struct F { void operator () () {} } it seems to be impossible to convert &F::operator () to instance of type using P = void (*)();.
Code:
struct L
{
static
void operator () () const {}
operator auto () const
{
return &L::operator ();
}
};
The error is
overloaded 'operator()' cannot be a static member function
but operator () is not overloaded.
Per standard 13.5/6,
An operator function shall either be a non-static member function or be a non-member function and have
at least one parameter whose type is a class, a reference to a class, an enumeration, or a reference to an
enumeration.
Additionally, in 13.5.4 it is stated that
operator()
shall be a non-static member function with an arbitrary number of parameters. It can have
default arguments. It implements the function call syntax
postfix-expression
(
expression-list
opt
)
where the
postfix-expression
evaluates to a class object and the possibly empty
expression-list
matches
the parameter list of an
operator()
member function of the class. Thus, a call
x(arg1,...)
is interpreted
as
x.operator()(arg1, ...)
for a class object
x
of type
T
I would think that there's no technical reason to forbid this (but not being familiar with the de-facto cross-vendor C++ ABI (Itanium ABI), I can't promise anything).
There's however an evolutional issue about this at https://cplusplus.github.io/EWG/ewg-active.html#88 . It even has the [tiny] mark on it, making it a somewhat "trivial" feature under consideration.
I can't see any technical reason to forbid a static auto operator()( ... ). But it's a special case, so it would complicate the standard to add support for it. And such complication is not necessary, because it's very easy to emulate:
struct L
{
static void func() {}
void operator()() const { func(); }
operator auto () const
{
return &L::func;
}
};
See Johannes' answer for some possibly useful extra info.
Like the others, I don't see a fundamental reason why it is not possible to have a static operator(), for stateless functors or in general.
(EDIT 2020: Just found this proposal http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2018/p1169r0.html)
(UPDATE 2021: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2021/p1169r1.html)
In some cases might conflict, according to other rules, with member/static overloading which is not allowed in C++ (again, not sure why).
struct A{
void f();
static int f(); // compile error
}
So even if it were allowed to have a static operator(), should this be permitted?
struct A{
void operator()();
static int operator()(); // should be a compiler error???
}
Anyway, there is only one true reason to have a static operator() that it is not purely a syntactic reason and it is that objects should be able to call static functions as if they were member functions.
struct A{
static int f():
}
...
A a;
a.f(); // calls A::f() !!!
Specifically, the user of the class A doesn't need to know if a function is implemented as static or as a member.
It can later be upgraded to a member function from a generic point of view.
Leaving that important application to generic programming aside, there is a workaround the leads to a similar syntax that I saw in https://quuxplusone.github.io/blog/2018/03/19/customization-points-for-functions/, and that is to have a static member function called _, a name that doesn't imply any meaning.
struct A{
static int _();
}
...
A::_(); // instead of the more desirable (?) A::() or A::operator()
a._(); // this invokes the static functon _
Instead of the more desirable A::() or A::operator(), (well are they desirable at all? I don't know; something like A() would be really interesting but doens't even follow the syntax of a static function and can be confused with a constructor).
(As I said, the only feature I still miss, regarding this limitation you point out, is that a() cannot automatically delegate to a static version of the operator(), e.g. A::operator().)
In summary, your code could look like this:
struct L{
static void _() {}
auto operator()() const{
return L::_();
}
};
L ell;
ell(); // calls L::_() really.
Not sure what are willing to achieve still.
One can "CRTP" the idea https://godbolt.org/z/74vxsTYxd
#include<utility> // forward
template<class Self>
struct stateless_functor_facade{
template<class... Args>
constexpr decltype(auto) operator()(Args&&... args) const{
return Self::operator_paren(std::forward<Args>(args)...);
}
};
struct square : stateless_functor_facade<square>{
static auto operator_paren(double x){return x*x;}
};
int main(){
square s;
s(5);
}
Starting with C++23 both operator() (P1169) and operator[] (P2589) can be static. There was never really a good reason for the restriction, so it's finally being lifted.
You can also define a Lambda as static by [] static { ... }. If you have a Lambda without captures, you always should define it as static.
Unfortunately, this behavior cannot become the implicit standard without breaking API and ABI compatibility, so the explicit definition is necessary. Backwards compatibility is, as so often, both a blessing and a curse.
i i'm not going to pretend to be an expert at all. however, i was looking into static lamdas or static operator() and theirs an ongoing proposal about it. i was satisfied with what i could understand.
it appears the ideas proposed is grappling with how to design it to not break other code.etc. however it appears their working on a solution. so. mabyeeee one day!
static operator()
Document #: P1169R2
Date: 2021-08-14
Project: Programming Language C++
3 Proposal
The proposal is to just allow the ability to make the call operator a
static member function, instead of requiring it to be a non-static
member function. We have many years of experience with member-less
function objects being useful. Let’s remove the unnecessary object
parameter overhead. There does not seem to be any value provided by
this restriction.
There are other operators that are currently required to be
implemented as non-static member functions - all the unary operators,
assignment, subscripting, conversion functions, and class member
access. We do not believe that being able to declare any of these as
static will have as much value, so we are not pursuing those at this
time. We’re not aware of any use-case for making any of these other
operators static, while the use-case of having stateless function
objects is extremely common.
find full Document here:
http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2021/p1169r2.html
It is now possible to have static operator (), after c++23 - go to https://en.cppreference.com/w/cpp/feature_test#Language_features and look for __cpp_static_call_operator. GCC 13 will supports it - https://gcc.gnu.org/projects/cxx-status.html.
Here is an example on Compiler Explorer with gcc trunk.
struct A {
constexpr static int operator() (int x) noexcept { return x - 1; }
constexpr static int operator[] (int x) noexcept { return x - 1; }
};
int main() {
return [](int x) constexpr static noexcept { return x - 1; }( A()( A()[3] ));
}
A simple, a little bit dirty workaround until the relevant committee considers this trivial feature:
Glob operators are syntactically similar to the constructors.
Thus, you can't write a
static MyClass::operator()(...);
It was made simply impossible, because a committee decided so on unclear reasons. I would be so happy if I could talk with one of their members, to ask, what was in their mind as they decided so. Unfortunately, he probably wouldn't understand my question, because he never programmed c++. He only worked on its docs.
Now they need some decades of
debates
consultations
meetings
and considerations
to implement a trivial feature. I suspect the feature may be made available around in c++3x, and on emulated machines it may be even tried out.
Until then, you can try to write:
MyClass::MyClass(...);
In both cases you can call MyClass(...);.
Of course it is useful mainly if MyClass is a singleton. And, I would say it is a hack. Furthermore, it allocates a sizeof(MyClass) on the stack which may be bad in the performance / efficiency side.
Furthermore, this will be in essence a constructor, and constructors can't return anything. But you can avoid this by storing the result in the instance, and then casting it by a cast operator to anything you wish to.
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
}
can someone tell me what is the different between (*ptr).field and ptr->field?
I know it connect somehow to static and dynamic linking, but i dont know what is it.
can someone tell me the differnet and give me an example?
edit:
if i have this code:
Point p; //point is a class that derive from class shape
Shape *s=&p;
//there is a diffrence if i write:
(*s).print(); //print is virtual func
s->print(); // the answers will not be the same, why?
TNX!
This has nothing to do with static or dynamic linking.
See C++'s operator precedence. The . has a lower precedence than *, so there's actually quite a difference between *ptr.fld and ptr->fld. For example, the following code demonstrates:
#include <iostream>
struct foo {
int f;
};
int main() {
struct foo *p = new struct foo;
p->f = 42;
std::cout << p->f << std::endl;
std::cout << (*p).f << std::endl;
// The following will not compile
// std::cout << *p.f << std::endl;
}
As John Knoeller points out, ptr->fld is syntactic sugar for (*(ptr)).fld, but is not the same as *ptr.fld, which would actually evaluate to *(ptr.fld), probably not what you want.
You'd use ptr->fld when you have a pointer to a structure and want to access a field contained therein. (*(ptr)).fld means the same thing but is not as tidy. You'd use *strct.fld when you have a structure, not a pointer to a structure, that contains a field (fld) which is a pointer you want to dereference. The case of ptr->fld is shown above. The case of *strct.fld could be used with the following structure:
struct foo {
int *fld;
}
struct foo f;
f.fld = new int;
*f.fld = 42;
it has nothing to do with static or dynamic linking
both expressions will return the value of ptr.field
the ptr->field form is an abbreviated syntax for accessing a member directly from a pointer
UPDATE: it occurred to me that your original intent was not linking but binding
if this indeed was what you were aiming to then there is static binding and dynamic binding
which have some relation to the -> operator see here
Static linking is the result of the linker copying all library routines used in the program into the executable image. This may require more disk space and memory than dynamic linking, but is both faster and more portable, since it does not require the presence of the library on the system where it is run.
Dynamic linking is accomplished by placing the name of a sharable library in the executable image. Actual linking with the library routines does not occur until the image is run, when both the executable and the library are placed in memory. An advantage of dynamic linking is that multiple programs can share a single copy of the library.
But it is unrelated to pointer indirection you've mentioned -- in fact, those two expressions are identical.
I assume that by *ptr.field you meant (*ptr).field.
As far as only built-in operators are considered, there's no difference between the two. And no, it has nothing to do with and "static" or "dynamic" linking, whatever is implied by these terms.
The only potential difference between the two is that in ptr->field variant -> is an overloadable operator in C++, while in the (*ptr).field variant only * is overloadable, while . is not.
Also, some differences between these two methods of member access existed in very archaic versions of C language (CRM C), but I doubt anyone cares about those today.
This has nothing to do with linking.
ptr->fld
is merely shorthand for
(*ptr).fld
It's pure syntactic sugar ;)
As long as ptr is a pointer, the two are equivalent once properly parenthesized (as others have said). If ptr is an object, rather than a pointer, they might be different depending on definitions for operator* and operator-> (if any) from the object's class or ancestors.
The form -> is just shorthand for de-refrencing the pointer and accessing the member.
(*ptr).field;
// Equiv to
ptr->field;
One good reason for using -> is that when you are following a chain:
int x = (*(*(*(*ptr).field1).field2).field3).field4;
// Equiv to
int y = ptr->field1->field2->field3->field4;
The second one becomes much more readable.
As for the second part of your question.
I find it really easy to just nock up an example.
#include <iostream>
class Shape
{
public: virtual ~Shape() {}
virtual void Print() {std::cout << "Shape\n";}
};
class Point: public Shape
{
public: virtual void Print() {std::cout << "Point\n";}
};
int main ()
{
Point p;
Shape* s = &p;
s->Print();
(*s).Print();
}
> vi x.cpp
> g++ x.cpp
> ./a.exe
Point
Point
As you can see the result is the same in both situations.
When you call a method via a pointer or a reference the virtual call mechanism will be invoked. The star operator (AKA derefence operator) returns a reference to an object (it does not actually de-reference the object). So when it is used to call a method the virtual call mechanism will be invoked and the most derived version of the method called.
They are both the same.
*ptr.field dereferences the variable ptr then returns the value of member field.
The -> operator is shorthand notation for the above.
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.
Closed 11 years ago.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
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.