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What is the meaning of “operator bool() const”

Posted by: admin November 30, 2017 Leave a comment

Questions:

For example:

operator bool() const 
{ 
    return col != 0; 
}

col is an int.
How does operator bool() const work?

Answers:

Member functions of the form

operator TypeName()

are conversion operators. They allow objects of the class type to be used as if they were of type TypeName and when they are, they are converted to TypeName using the conversion function.

In this particular case, operator bool() allows an object of the class type to be used as if it were a bool. For example, if you have an object of the class type named obj, you can use it as

if (obj)

This will call the operator bool(), return the result, and use the result as the condition of the if.

It should be noted that operator bool() is A Very Bad Idea and you should really never use it. For a detailed explanation as to why it is bad and for the solution to the problem, see “The Safe Bool Idiom.”

(C++0x, the forthcoming revision of the C++ Standard, adds support for explicit conversion operators. These will allow you to write a safe explicit operator bool() that works correctly without having to jump through the hoops of implementing the Safe Bool Idiom.)

Questions:
Answers:
operator bool() const 
{
    return col != 0;
}

Defines how the class is convertable to a boolean value, the const after the () is used to indicate this method does not mutate (change the members of this class).

You would usually use such operators as follows:

airplaysdk sdkInstance;
if (sdkInstance) {
    std::cout << "Instance is active" << std::endl;
} else {
    std::cout << "Instance is in-active error!" << std::endl;
}

Questions:
Answers:

I’d like to give more codes to make it clear.

struct A
{
    operator bool() const { return true; }
};

struct B
{
    explicit operator bool() const { return true; }
};

int main()
{
    A a1;
    if (a1) cout << "true" << endl; // OK: A::operator bool()
    bool na1 = a1; // OK: copy-initialization selects A::operator bool()
    bool na2 = static_cast<bool>(a1); // OK: static_cast performs direct-initialization

    B b1;     
    if (b1) cout << "true" << endl; // OK: B::operator bool()
    // bool nb1 = b1; // error: copy-initialization does not consider B::operator bool()
    bool nb2 = static_cast<bool>(b1); // OK: static_cast performs direct-initialization
}

Questions:
Answers:

It’s user-defined implicit conversion function to convert your class into either true or false.

//usage
bool value = yourclassinstance; //yourclassinstance is converted into bool!

Questions:
Answers:

It’s an implicit conversion to bool. I.e. wherever implicit conversions are allowed, your class can be converted to bool by calling that method.

Questions:
Answers:

As the others have said, it’s for type conversion, in this case to a bool. For example:

class A {
    bool isItSafe;

public:
    operator bool() const
    {
        return isItSafe;
    }

    ...
};

Now I can use an object of this class as if it’s a boolean:

A a;
...
if (a) {
    ....
}

Questions:
Answers:

Another common use is for std containers to do equality comparison on key values inside custom objects

class Foo
{
    public: int val;
};

class Comparer { public:
bool operator () (Foo& a, Foo&b) const {
return a.val == b.val; 
};

class Blah
{
std::set< Foo, Comparer > _mySet;
};