I have some special numeric class. The class can be constructed by a double:
struct FancyDouble{
FancyDouble& operator = (double v){
this->value = v;
return *this;
}
};
FancyDouble myFancyDouble = 5.0;
The class provides an (implicit) conversion from double to FancyDouble. Is there a way to do the opposite? So when I use a FancyDouble in a numeric expression, I'd like to write:
double a = 123;
double b = b * myFancyDouble;
I had a look into the reference and I think it's not possible, but I'm not absolutely sure.
Yes, you can. They are called conversion operators:
struct FancyDouble
{
// your things here
operator double() { return this->value; }
};
You can have a conversion to any type, not just built-in types. They are also typically marked const (though I didn't do it here to show the core concept).
Note that your struct (as shown) doesn't exactly allow implicit conversions from double to FancyDouble; your operator= only means that you can use a double on the right-hand side of an assignment, but if you have a FancyDouble function parameter, it won't work since what you would need there is a constructor that accepts a double, not an assignment operator.
void foo(FancyDouble fd);
foo(4.4); // does not work because you can't construct a FancyDouble from a double
// even though you have an assignment operator
You would do something like this:
struct FancyDouble
{
// your things here
FancyDouble(double d) { /* initialization here */ }
};
With this, you would only need an operator= that accepts a FancyDouble instead of a double, and the conversion would be done automatically:
struct FancyDouble
{
// your things here
FancyDouble(double d) { /* initialization here */ }
FancyDouble& operator=(FancyDouble fd) { /* assignment here */ }
};
FancyDouble fd = 4.2; // works
fd = 5.5; // also works, because FancyDouble can be constructed implicitly from a double
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.
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;
}
With all of the fundamental types of C++, one can simply query:
if(varname)
and the type is converted to a boolean for evaluation. Is there any way to replicate this functionality in a user-defined class? One of my classes is identified by an integer, although it has a number of other members, and I'd like to be able to check if the integer is set to NULL in such a manner.
Thanks.
The C++11 approach is:
struct Testable
{
explicit operator bool() const
{ return false; }
};
int main ()
{
Testable a, b;
if (a) { /* do something */ } // this is correct
if (a == b) { /* do something */ } // compiler error
}
Note the explicit keyword which prevents the compiler from converting implicitly.
You can define a user-defined conversion operator. This must be a member function, e.g.:
class MyClass {
operator int() const
{ return your_number; }
// other fields
};
You can also implement operator bool. However, I would STRONGLY suggest against defining conversion operators to integer types (including bool) because your class will become usable in arithmetic expressions which can quickly lead to a mess.
As an alternative, for example, IOStreams define conversion to void*. You can test void* in the same way you can test a bool, but there are no language-defined implicit conversions from void*. Another alternative is to define operator! with the desired semantics.
In short: defining conversion operators to integer types (including booleans) is a REALLY bad idea.
Simply implement operator bool() for your class.
e.g.
class Foo
{
public:
Foo(int x) : m_x(x) { }
operator bool() const { return (0 != m_x); }
private:
int m_x;
}
Foo a(1);
if (a) { // evaluates true
// ...
}
Foo b(-1);
if (b) { // evaluates true
// ...
}
Foo c(0);
if (c) { // evaluates false
// ...
}
As others have stated, using operator int () or operator bool () is bad idea because of the conversions it allows. Using a pointer is better idea. The best know solution to this problem so far is to return a member (function) pointer:
class MyClass {
void some_function () {}
typedef void (MyClass:: * safe_bool_type) ();
operator safe_bool_type () const
{ return cond ? &MyClass::some_function : 0; }
};
C++ checks if the statements result is whether equal to zero nor not. So i think you can define equality operator for your class and define how your class will be different from zero in which conditions.
Okay so I have a class that has 'weak typing' I.E. it can store many different types defined as:
#include <string>
class myObject{
public:
bool isString;
std::string strVal;
bool isNumber;
double numVal;
bool isBoolean;
bool boolVal;
double operator= (const myObject &);
};
I would like to overload the assignment operator like this:
double myObject::operator= (const myObject &right){
if(right.isNumber){
return right.numVal;
}else{
// Arbitrary Throw.
throw 5;
}
}
So that I can do this:
int main(){
myObject obj;
obj.isNumber = true;
obj.numVal = 17.5;
//This is what I would like to do
double number = obj;
}
But when I do that, I get:
error: cannot convert ‘myObject’ to ‘double’ in initialization
At the assignment.
I have also tried:
int main(){
myObject obj;
obj.isNumber = true;
obj.numVal = 17.5;
//This is what I would like to do
double number;
number = obj;
}
To which I get:
error: cannot convert ‘myObject’ to ‘double’ in assignment
Is there something I am missing? or is it simply not possible to do a conversion like that by overloading operator=.
Overloading operator= changes the behaviour when assigning to objects of your class type.
If you want to provide implicit conversion to other types you need to supply a conversion operator, e.g.
operator double() const
{
if (!isNumber)
throw something();
return numVal;
}
What you really want are conversion operators.
operator double() const { return numVal; }
operator int() const { ...
That said, you'd probably like boost::variant.
for that to work, you need to implement a conversion operator from your object to something that can be converted into a double
The return value of operator=() cannot be used as you have tried to demonstrate. If you think of the overloaded operator as a function in it's own right, it may make more sense.
For example:
int main() {
myObject obj, obj2;
obj.isNumber = true;
obj.numVal = 17.5;
obj2.operator=(obj); // equivalent to obj2 = obj
}
The reason number = obj; doesn't work is because you've defined myObject::operator=(), whereas number would be using double::operator=() (okay, technically there is no double::operator=() since it's a fundamental type and not a class...just work with me here).
An interesting note is that this function behaves like any other function in that the return value (return right.numval;) is ignored when it's not used. However, the return value can be assigned or used like the return value of any other function, so if you really wanted you could do something like this:
int main() {
myObject obj, obj2;
obj.isNumber = true;
obj.numVal = 17.5;
double number;
// number = obj; still won't work.
number = obj2 = obj; // equivalent to number = obj2.operator=(obj)
}
This is only so useful. As others have mentioned, you really want to look into conversion operators when trying to assign myObject objects to fundamental types.
To make a class assignable to a double, operator= must be defined differently.
double operator=(myclass&) is wrong.
What would work, is a friend operator= outside your class, which accepts double and myclass&.