How would I overload the = operator in a way that I could execute
int someInt;
MyClass instanceOfMyClass;
someInt = instanceOfMyClass;
Where MyClass contains an integer named number?
You can't overload the operator= for the type int. What you are really looking for is the conversion operator operator int() for your MyClass. In your case, considering x to be the private member of your class:
operator int() { return x; }
I would avoid doing this in most cases as it can be difficult to tell what exactly is happening. That said:
class MyClass {
public:
operator int() { return number; }
private:
int number;
};
This creates an (implicit) conversion operator for your class.
Related
i am trying to implement a "Integer" class with expect that it will work like a build-in type int. But i am having a problem: i can't use this class in operator [] like int See my code below:
#include <iostream>
using namespace std;
class Integer
{
private:
int Value;
public:
Integer(int x = 0)
{
Value = x;
}
void SetValue(int x)
{
Value = x;
}
int GetValue()
{
return Value;
}
Integer& operator [] (Integer X);
};
Integer& Integer::operator [] (Integer X)
{
// Code
}
int main()
{
Integer X[10];
Integer I(5);
int i = 5;
for(int i=0; i<10; ++i)
X[i].SetValue(i+1);
cout << X[i].GetValue() << endl; // It still work
cout << X[I].GetValue() << endl; // Doesn't work
return 0;
}
What is a way (exclude cast operator) to make operator [] understand my Integer type like it does with int?
You are thinking about this the wrong way around. You don't need to overload the [] operator on your own class, you actually need your class to be convertable to int, which can be done by overloading the cast operator.
class Integer
{
public:
operator int () const { return Value; }
};
There's a bit of confusion here. The declaration of an array of type T takes the form
T t[n];
where n is a compile time evaluable constant expression for the array size.
This does not invoke any [] operator defined within T, and neither does a subsequent access of an element via the expression t[i] for an integral type i.
If, in your case, you want X[I] to be compilable then you need to provide an operator that allows I to be treated as an array index (more formally the language requires a primitive integral type). A cast operator to int is the obvious choice, with
operator int() const
{
return Value;
}
being the implementation.
You'll need to add a type cast operator to your class. This allows your class to be converted to a compatible type. See here for more information: http://www.cplusplus.com/doc/tutorial/typecasting/
This function should do the trick for your class.
operator int() const { return GetValue(); }
Actually, you don't need to overload the [] operator.
You just need to make sure that you can convert your Integer into an int with int int operator.
Something like this would work better.
operator int() const
{ return GetValue(); }
Then you should also be carefull with your prints.
X[i].GetValue() is wrong since i = 10 in your exemple. This will result in an execution error.
X[I].GetValue() is wrong in your exemple because de conversion between your class and an int is not possible without I.getValue(). I am surprise this doesn't result in a compilation error but overloading the int operator will result this issue tho.
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'm fairly familiar with operator overloading, however I am wondering how do we implement something like this:
myClass myclassobj;
int x;
x = 5;
x = x + myclassobj
There is no way to overload the + operator for the int class, so something should be done from myClass, but how would we do that? I am probably using the wrong keywords, but searching through SO didn't lead to anithing. Apologies if I did something wrong, this is my first post here.
Edit - My class is a custom vector class, so simply converting it to given type won't work.
Define an overloaded operator with the signature int operator+(int, myClass).
You are right.
There is no way to overload the + operator for the int class, so
something should be done from myClass
Your question:
but how would we do that?
My answer:
You should use user defined type conversion. It may work with a conversion operator.
#include <iostream>
class myClass
{
int i;
public:
myClass(int i=0) : i(i) { }
operator int(){ // A conversion from myClass to int may solve your problem.
return i;
}
};
int main()
{
myClass myclassobj(99);
int x=7;
x = 5;
x = x + myclassobj;
std::cout<<x<<std::endl;
return 0;
}
Brian gave a good answer also, but it works only if the overloaded operator does not need protected or private members from the second argument or if overloaded operator declared as friend of myClass.
The title basically says it all. I mainly want to do this so that I can create an object (say, a custom string object) that can initialize the parameters of other functions in other APIs. Here's an example of me trying to get a custom integer class to work:
#include <iostream>
using namespace std;
class test
{
public:
int member;
test(int i) : member(i) {}
friend int &operator=(int &i, test t);
};
int &operator=(int &i, test t)
{
return (i = t.member);
}
int main()
{
int i;
test t = 90;
cout << (i = t);
return 0;
}
Unfortunately I receive an error saying that operator= needs to be a member function. I understand the C++ standard's goal in preventing static and non-member overloads for the assignment operator from being implemented, but is there any other way to do this? Thanks for any help/suggestions!
This is not done with an assignment operator but with an overloaded typecast. This would make your main function work like expected:
#include <iostream>
using namespace std;
class test
{
public:
int member;
test(int i) : member(i) {}
operator int() const {return member;}
};
int main()
{
int i;
test t = 90;
cout << (i = t);
return 0;
}
What you are trying to do needs an conversion operator
operator int()
{
return this->member;
}
For the class you are trying to write(containing only integer members), You do not need to overload the = operator.
= operator is one of the member functions that is generated by the compiler by default for every class. Caveat is, it does a simple bit by bit copy(shallow copy) of class members, since you have only integers it should be good enough for you.
You would need to overload the = operator if you had dynamically allocated pointers as member functions, because in that case a shallow copy of those pointers would result in all the objects containing a member pointer pointing to the same dynamic memory location & if one of the object finishes it lifetime, other objects are left with a dangling pointer.
As #Tony, aptly points in out comments Shallow copy is usually bad but not always. See his comments for a scenario.
If at all you want to overload the assignment operator check out the Copy and Swap Idiom to do it right way.
You should also check out the Rule of Three.
Try this:
class test
{
public:
int member;
test(int i) : member(i) {}
operator int() {return this->member;}
};
int main(void)
{
int i;
test t = 90;
cout << (i = t);
return 0;
}
The assignment operator cannot be a friend function. The assignment operator can only be declared as a non-static member function. This is to ensure that it receives the L-value as its first operand. The same is true for the [], (), and -> operators. In your case, since int is an build-in type, you cannot use member function. You can implement operator int() to cast your user-defined type to int.
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&.