Why can I assign struct with a pointer?

Question

Why does it even compile?

struct UE{
    UE(bool a = true) {  };
//  UE(){}; // if UE took no initial args and called below, gcc will complain.
};
class VA {
    protected:
            UE ue;
    public:
            VA();
};
VA::VA()
{
    ue = new UE(true); // ???why???
//  ue = new UE(); // will complain
}

I tried with gcc (GCC) 4.6.2. How can I assign a struct with a pointer?

Answer 1

You're hitting the fact that you didn't mark your constructor as explicit , and thus it can be used for implicit conversions.

new UE(true) returns a pointer. All pointers can be implicitly converted to bool , resulting in true if they are non-null. UE can be implicitly constructed from a bool . So in effect, the pointer returned by new is converted to bool , which is converted to UE using your constructor, and then the copy assignment operator of UE is called. Of course, the memory allocated by new is leaked.

The take-home message is: always mark your single-argument constructors as explicit unless you actually want them to be usable for implicit conversions. By "single-argument constructor," I mean one which can be called with a single argument. Either because it has one parameter, or it has more and all parameters after the first have default arguments.

Answer 2

There is an implicit conversion from any pointer to bool . Hence this code is implicitly calling the single argument constructor after converting the pointer to a bool value

One way to fix this is to make the single parameter constructor explicit.

struct UE{
    explicit UE(bool a = true) {  };
};

This will prevent the code from compiling unless the constructor is explicitly invoked

Answer 3

Any scalar type can be converted to boolean value true or false depending on whether its value is equal to zero or not..

In this statement

ue = new UE(true); 

the implicitly defined by the compiler copy assignment operator is called. As expression new UE( true ) is not equal to zero then it can be converted implicitly to boolean value true. Class UE has conversion constructor

UE(bool a = true);

that converts an object of type bool to an object of type UE.

To prevent such usage you should define the constructor as explicit

explicit UE(bool a = true);

Answer 4

If you mark you constructor explicit this won't work:

explicit UE( bool a = true) {  };

this is because there is an implicit conversion from the pointer to a bool, we can see this is allowed from draft C++ standard section 4.12 Boolean conversions says ( emphasis mine ):

A prvalue of arithmetic, unscoped enumeration, pointer , or pointer to member type can be converted to a prvalue of type bool.