Is it possible to change the behavior of if() so that:
class Foo {
int x;
};
Foo foo;
if(foo)
only proceeds if the value of x is something other than zero? or...
Would an explicit user-defined type conversion to int work/would that be an appropriate approach? or...
Is it best to do something like if(foo.getX())?
You can convert your object to a boolean value by defining operator bool():
explicit operator bool() const
{
return foo.getX();
}
The explicit keyword prevents implicit conversions from Foo to bool. For example, if you accidentally put foo in an arithmetic expression like foo + 1, the compiler could detect this error if you declare operator bool() as explicit, otherwise foo will be converted to bool even if not intended.
In general, member functions of the form
operator TypeName()
(with optional explicit and const qualifier) are conversion operators. It allows you to cast your class to any type specified by TypeName. In the other direction, constructors with one argument allow you to cast any type to your class:
class Foo {
Foo(int x); // convert int to Foo
operator bool() const; // convert Foo to bool
int x;
};
This defines implicit conversions for your class. The compiler tries to apply these conversions if possible (like what it does for built-in data types, e.g. 5 + 1.0). You can declare them to be explicit to suppress unwanted implicit conversions.
You can define an operator to convert the object to bool
class Foo
{
int x;
public:
operator bool() const
{
return x > 0;
}
};
But this can have unintended consequences because of implicit conversions to bool when you don't desire the conversion to take place. For instance
int x = 42 + Foo();
C++11 solves this problem by allowing you to declare the conversion operator as explicit, which then only allows implicit conversions in certain contexts, such as within an if statement.
explicit operator bool() const // allowed in C++11
Now
int x = 42 + Foo(); // error, no implicit conversion to bool
int x = 42 + static_cast<bool>(Foo()); // OK, explicit conversion is allowed
Related
Consider this simple example:
template <class Type>
class smartref {
public:
smartref() : data(new Type) { }
operator Type&(){ return *data; }
private:
Type* data;
};
class person {
public:
void think() { std::cout << "I am thinking"; }
};
int main() {
smartref<person> p;
p.think(); // why does not the compiler try substituting Type&?
}
How do conversion operators work in C++? (i.e) when does the compiler try substituting the type defined after the conversion operator?
Some random situations where conversion functions are used and not used follow.
First, note that conversion functions are never used to convert to the same class type or to a base class type.
Conversion during argument passing
Conversion during argument passing will use the rules for copy initialization. These rules just consider any conversion function, disregarding of whether converting to a reference or not.
struct B { };
struct A {
operator B() { return B(); }
};
void f(B);
int main() { f(A()); } // called!
Argument passing is just one context of copy initialization. Another is the "pure" form using the copy initialization syntax
B b = A(); // called!
Conversion to reference
In the conditional operator, conversion to a reference type is possible, if the type converted to is an lvalue.
struct B { };
struct A {
operator B&() { static B b; return b; }
};
int main() { B b; 0 ? b : A(); } // called!
Another conversion to reference is when you bind a reference, directly
struct B { };
struct A {
operator B&() { static B b; return b; }
};
B &b = A(); // called!
Conversion to function pointers
You may have a conversion function to a function pointer or reference, and when a call is made, then it might be used.
typedef void (*fPtr)(int);
void foo(int a);
struct test {
operator fPtr() { return foo; }
};
int main() {
test t; t(10); // called!
}
This thing can actually become quite useful sometimes.
Conversion to non class types
The implicit conversions that happen always and everywhere can use user defined conversions too. You may define a conversion function that returns a boolean value
struct test {
operator bool() { return true; }
};
int main() {
test t;
if(t) { ... }
}
(The conversion to bool in this case can be made safer by the safe-bool idiom, to forbid conversions to other integer types.) The conversions are triggered anywhere where a built-in operator expects a certain type. Conversions may get into the way, though.
struct test {
void operator[](unsigned int) { }
operator char *() { static char c; return &c; }
};
int main() {
test t; t[0]; // ambiguous
}
// (t).operator[] (unsigned int) : member
// operator[](T *, std::ptrdiff_t) : built-in
The call can be ambiguous, because for the member, the second parameter needs a conversion, and for the built-in operator, the first needs a user defined conversion. The other two parameters match perfectly respectively. The call can be non-ambiguous in some cases (ptrdiff_t needs be different from int then).
Conversion function template
Templates allow some nice things, but better be very cautious about them. The following makes a type convertible to any pointer type (member pointers aren't seen as "pointer types").
struct test {
template<typename T>
operator T*() { return 0; }
};
void *pv = test();
bool *pb = test();
The "." operator is not overloadable in C++. And whenever you say x.y, no conversion will automatically be be performed on x.
Conversions aren't magic. Just because A has a conversion to B and B has a foo method doesn't mean that a.foo() will call B::foo().
The compiler tries to use a conversion in four situations
You explicitly cast a variable to another type
You pass the variable as an argument to a function that expects a different type in that position (operators count as functions here)
You assign the variable to a variable of a different type
You use the variable copy-construct or initialize a variable of a different type
There are three types of conversions, other than those involved with inheritance
Built-in conversions (e.g. int-to-double)
Implicit construction, where class B defines a constructor taking a single argument of type A, and does not mark it with the "explicit" keyword
User-defined conversion operators, where class A defines an operator B (as in your example)
How the compiler decides which type of conversion to use and when (especially when there are multiple choices) is pretty involved, and I'd do a bad job of trying to condense it into an answer on SO. Section 12.3 of the C++ standard discusses implicit construction and user-defined conversion operators.
(There may be some conversion situations or methods that I haven't thought of, so please comment or edit them if you see something missing)
Implicit conversion (whether by conversion operators or non-explicit constructors) occurs when passing parameters to functions (including overloaded and default operators for classes). In addition to this, there are some implicit conversions performed on arithmetic types (so adding a char and a long results in the addition of two longs, with a long result).
Implicit conversion does not apply to the object on which a member function call is made: for the purposes of implicit conversion, "this" is not a function parameter.
The compiler will attempt one(!) user-defined cast (implicit ctor or cast operator) if you try to use an object (reference) of type T where U is required.
The . operator, however, will always try to access a member of the object (reference) on its left side. That's just the way it's defined. If you want something more fancy, that's what operator->() can be overloaded for.
You should do
((person)p).think();
The compiler doesn't have the information for automatically casting to person, so you need explicit casting.
If you would use something like
person pers = p;
Then the compiler has information for implicit casting to person.
You can have "casting" through constructors:
class A
{
public:
A( int );
};
A a = 10; // Looks like a cast from int to A
These are some brief examples. Casting (implicit, explicit, etc) needs more to explain. You can find details in serious C++ books (see the questions about C++ books on stack overflow for good titles, like this one).
//Virtual table Fuction(VFT)
#include <iostream>
using namespace std;
class smartref {
public:
virtual char think() { }//for Late bindig make virtual function if not make virtual function of char think() {} then become early binding and pointer call this class function
smartref() : data(new char) { }
operator char(){ return *data; }
private:
char* data;
};
class person:public smartref
{
public:
char think() { std::cout << "I am thinking"; }
};
int main() {
smartref *p;//make pointer of class
person o1;//make object of class
p=&o1;//store object address in pointer
p->think(); // Late Binding in class person
return 0;
}
In the class below,
Why would you make the operators explicit. I thought that explicit was to prevent implicit calling of constructors?
class Content
{
public:
virtual ~Content() = 0;
virtual explicit operator float&();
virtual explicit operator long long&();
virtual explicit operator std::string&()
}
I thought that explicit was to prevent implicit calling of
constructors?
Since C++11 it also applies to user-defined conversions (a.k.a. the cast operator).
Why would you make the operators explicit
Used in this context, the explicit keyword makes the conversion eligible only for direct-initialization and explicit conversions. See here under [class.conv.fct¶2]:
A conversion function may be explicit ([dcl.fct.spec]), in which case
it is only considered as a user-defined conversion for
direct-initialization ([dcl.init]). Otherwise, user-defined
conversions are not restricted to use in assignments and
initializations.
This aids you in making sure the compiler doesn't try the conversion against your intention, so that you have to explicitly cast it yourself, leaving less room for error. Example:
struct Foo
{
explicit operator int() {return 0;}
operator int*() {return nullptr;}
};
int main()
{
Foo foo;
//int xi = foo; // Error, conversion must be explicit
int i = static_cast<int>(foo); // OK, conversion is explicit
int* i_ptr = foo; // OK, implicit conversion to `int*` is allowed
int i_direct(foo); // OK, direct initialization allowed
int* i_ptr_direct(foo); // OK, direct initialization after implicit conversion
return 0;
}
It can also help resolve ambiguity in cases where multiple conversion options apply, leaving the compiler without a criteria for deciding which one to choose:
struct Bar
{
operator int() {return 1;}
operator char() {return '1';}
};
int main()
{
Bar bar;
//double d = bar; // Error, implicit conversion is ambiguous
return 0;
}
Add explicit:
struct Bar
{
operator int() {return 1;}
explicit operator char() {return '1';}
};
int main()
{
Bar bar;
double d = bar; // OK, implicit conversion to `int` is the only option
return 0;
}
Consider the following:
struct Content
{
operator float() { return 42.f; }
friend Content operator+(Content& lhs, float) { return lhs; }
};
int main()
{
Content c{};
c + 0; // error: ambiguous overload for 'operator+'
}
Here, the compiler cannot choose between operator+(Content&, float) and operator+(float, int). Making the float operator explicit resolves this ambiguity*:
c + 0; // operator+(Content&, float)
or
static_cast<float>(c) + 0; // operator+(float, int)
*) provided it makes sense to prefer one over the other.
The other answers cover how it works, but I think you should be told why it was added to C++.
A smart pointer usually has a conversion to bool so you can do this:
std::shared_ptr<int> foo;
if (foo) {
*foo = 7;
}
where if(foo) converts foo to bool. Unfortunately:
int x = foo+2;
converts foo to bool, then to int, then adds 2. This is almost always a bug. This is permitted because while only one user defined conversion is done, a user defined conversion followed by a built in conversion can silently occur.
To fix this programmers would do crazy things like add:
struct secret {
void unused();
};
struct smart_ptr {
using secret_mem_ptr = void(secret::*)();
operator secret_mem_ptr() const { return 0; }
};
and secret_mem_ptr is a secret pointer to member. A pointer to member has a built in conversion to bool, so:
smart_ptr ptr;
if (!ptr) {
}
"works" -- ptr is convert to secret_mem_ptr, which then is convered to bool, which is then used to decide which branch to take.
This was more than a bit of a hack.
They added explicit on conversion operators to solve this exact problem.
Now:
struct smart_ptr {
explicit operator bool() const { return true; }
};
doesn't permit:
smart_ptr ptr;
int x = 3 + ptr;
but it does permit:
if (ptr) {
}
because the rules were hand-crafted to support exactly that use case. It also doesn't permit:
bool test() {
smart_ptr ptr;
return ptr;
}
here, you have to type:
bool test() {
smart_ptr ptr;
return (bool)ptr;
}
where you explicitly convert ptr to bool.
(I am usually really against C-style casts; I make an exception in the case of (bool)).
You would use it if you wanted a Content object never to be implicitly converted to (say) a float. This could happen in the following way:
void f( float f );
....
Content c;
f( c ); // conversion from Content to float
Without the explicit qualifier, the conversion happens implicitly; with it, you get a compilation error.
Silent, implicit conversions can be the source of a lot of confusion and/or bugs, so it's generally better to make the operators explicit , or probably better yet to provide named functions, such as ToFloat, which tell the reader exactly what is going on.
I am studying converting constructors and conversion operators in C++.
What I've learned so far is that any non-explicit constructor that takes only one parameter (and any number of optional default arguments) represents an implicit class-type conversion to THAT class type, for example if a class defines a constructor that has one parameter of type int I can use an int wherever an object of that class type is required:
(assuming class_type has an overloaded += operator)
class_type a;
a+=5;
in this case 5 is implicitly converted (through the converting constructor) to class_typeand the overloaded operator is called.
Now, the (at least for me) tricky part: I know I can define a conversion operator as a member function :
operator int() {....};
that converts the object of class_type to the primitive int type, and I can use that conversion like:
class_type a;
a+5;
in this case I've read that the object is converted to an int through its conversion operator and then the buil-in sum operator is called.
But what if I defined an overloaded + operator to take two class_type objects as its arguments? something like
class_type operator+(const class_type&,const class_type &c);
how is the compiler supposed to know which one to call through function matching?
does the conversion to int only happens implicitly when only the built-in operator is defined?
thanks!
edit:
actually,I've tried to write some code to effectively try it out, it turned out that my compiler (g++) doesn't issue any ambiguous call error!
this is the class header (along with the non-memeber operator+ function declaration) :
#include <iostream>
class wrapper {
friend std::ostream &operator<<(std::ostream&,const wrapper&);
public:
wrapper()=default;
wrapper(int);
int get();
operator int() const;
wrapper operator+(int);
private:
int a=10;
};
std::ostream &operator<<(std::ostream&,const wrapper&);
and this is the main code:
#include "wrapper.h"
int main()
{
using namespace std;
wrapper w1;
wrapper w2(5);
cout<<w1<<" "<<w2<<endl;
w1+1;
}
now,I've defined a converting constructor from int to wrapper AND a conversion operator from class type to int(I've also overloaded the << output operator in order to print some results), but when the compiler evaluates the expression w1+1 it seems to be fine. How could it possibly be??
If you have for example the following class declaration that contains a conversion constructor and a conversion operator
struct A
{
A( int x ) : x( x ) {}
operator int() const { return x; }
int x;
};
const A operator +( const A &a1, const A &a2 )
{
return A( a1.x + a2.x );
}
then statement
a1 + a2;
where a1 and a2 are declared like for example
A a1( 10 );
A a2( 20 );
will be well-formed because there is no need to call a conversion function. The both operands match the parameter declarations of the operator +.
However if you will write for example
a1 + 20;
when the compiler issues an error because there is an ambiguity. The compiler can either apply conversion constructor A( int ) to convert the second operand to type A and call the operator defined for objects of type A. Or it can apply the conversion operator operator int to convert the first operand to type int and call the built-in operator + for objects of type int.
To avoid this ambiguity you could declare either the constructor or the operator (or the both) with function specifier explicit.
For example
explicit A( int x ) : x( x ) {}
or
explicit operator int() const { return x; }
In this case only one implicit conversion would exist and there was not an ambigiuty.
I would like to append the above description that sometimes some converion operators can be called implicitly even if they are declared with the function specifier explicit.
For example According to the C++ Standard (6.4 Selection statements)
...The value of a condition that is an expression is the value of the
expression, contextually converted to bool for statements other
than switch;
and (5.16 Conditional operator)
1 Conditional expressions group right-to-left. The first expression is
contextually converted to bool (Clause 4).
So for example if the above class has the following conversion operator declared with the function specifier explicit
explicit operator bool() const { return x != 0; }
nevertheless it will be called implicitly for example in the following statement
A a( 10 );
std::cout << ( a ? "true" : "false" ) << std::endl;
Here a will be converted to an object of type bool in the conditional operator.
EDIT: After you updated your question this expression
w1+1;
is an exact match for operator
wrapper operator+(int);
Neither conversion are required. So the code compiles successfully.
This is something you can easily try and see what the compiler does:
#include <iostream>
struct ABC {
int v;
ABC(int x) : v(x) { }
operator int() const { return v; }
void operator +=(ABC const &that) {
v += that.v;
}
};
ABC operator+(ABC const &lhs, ABC const &rhs) {
return { lhs.v + rhs.v };
}
int main() {
ABC a(5);
std::cout << a + 1 << '\n';
a += 10;
std::cout << a << '\n';
}
what if I defined an overloaded + operator to take two class_type objects as its arguments?
GCC
error: ambiguous overload for 'operator+' (operand types are 'ABC' and 'int')
The compiler sees two candidates: operator+(int, int) <built-in> and ABC operator+(const ABC&, const ABC&). This means it could implicitly convert not only the 5 in a + 5 to a but also the a to int. Post these conversions both operator+ functions become potential matches.
How is the compiler supposed to know which one to call through function matching?
It doesn't know hence the error.
does the conversion to int only happens implicitly when only the built-in operator is defined?
Yes, otherwise it doesn't automatically convert class_type to int. However, int to class_type would happen implicitly unless you make class_type's constructor explicit:
explicit ABC(int x) : v(x) { }
If you've access to C++11, then you also make the conversion function explicit:
explicit operator int() const { return v; }
Suppose, if a conversion from one type another type is not available through explicit casts e.g static_cast, Would it be possible to define explicit conversion operators for it?
Edit:
I'm looking for a way to define explicit conversion operators for the following:
class SmallInt {
public:
// The Default Constructor
SmallInt(int i = 0): val(i) {
if (i < 0 || i > 255)
throw std::out_of_range("Bad SmallInt initializer");
}
// Conversion Operator
operator int() const {
return val;
}
private:
std::size_t val;
};
int main()
{
SmallInt si(100);
int i = si; // here, I want an explicit conversion.
}
For user defined types you can define a type cast operator. The syntax for the operator is
operator <return-type>()
You should also know that implicit type cast operators are generally frowned upon because they might allow the compiler too much leeway and cause unexpected behavior. Instead you should define to_someType() member functions in your class to perform the type conversion.
Not sure about this, but I believe C++0x allows you to specify a type cast is explicit to prevent implicit type conversions.
In the current standard, conversions from your type to a different type cannot be marked explicit, which make sense up to some extent: if you want to explicitly convert you can always provide a function that implements the conversion:
struct small_int {
int value();
};
small_int si(10);
int i = si.value(); // explicit in some sense, cannot be implicitly converted
Then again, it might not make that much sense, since in the upcoming standard, if your compiler supports it, you can tag the conversion operator as explicit:
struct small_int {
explicit operator int();
};
small_int si(10);
// int i = si; // error
int i = (int)si; // ok: explicit conversion
int j = static_cast<int>(si); // ok: explicit conversion
If this is what you want, you can define conversion operators, e.g.:
void foo (bool b) {}
struct S {
operator bool () {return true;} // convert to a bool
};
int main () {
S s;
foo (s); // call the operator bool.
}
Although it is not really recommended because once defined, such implicit conversion can occur in awkward places where you don't expect it.
Is there a way to define a type cast from say a user defined class to a primitive type (int, short, etc.)? Also, would any such mechanism require an explicit cast, or would it work implicitly?
for example:
// simplified example class
class MyNumberClass
{
private:
int value;
public:
// allows for implicit type casting/promotion from int to MyNumberClass
MyNumberClass(const int &v)
{
value = v;
}
};
// defined already above
MyNumberClass t = 5;
// What's the method definition required to overload this?
int b = t; // implicit cast, b=5.
// What's the method definition required to overload this?
int c = (int) t; // C-style explicit cast, c=5.
// ... etc. for other cast types such as dynamic_cast, const_cast, etc.
Yes, you can define an operator type() to do the conversion, and yes, it will work implicitly whenever such a conversion is required:
operator int() {
return value;
}