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())
?
NO. There is no such requirement that you Must overload != If you need to overload == . However,it is a good practice that you Should overload operators related to each other.
You cannot overload the conditional operator.
New operators can not be created. 2) Arity of the operators cannot be changed. 3) Precedence and associativity of the operators cannot be changed. 4) Overloaded operators cannot have default arguments except the function call operator () which can have default arguments.
Operator overloading is a compile-time polymorphism in which the operator is overloaded to provide the special meaning to the user-defined data type. Operator overloading is used to overload or redefines most of the operators available in C++. It is used to perform the operation on the user-defined data type.
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
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.
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