CODEWITHSUNDEEP
c++pointersstructreference
Beakie
Beakie

Reputation: 2009

Converting const struct reference to non-const pointer

I am trying to work out how to convert a const struct reference to a non-const struct pointer.

I have a struct called Foo:

struct Foo
{
    Foo& someFunc()
    {
        //do something

        return *this;
    }
};

I have a method:

struct Bar
{
    Foo* fooPointer;

    void anotherFunc(const Foo& foo)
    {
        // set fooPointer to foo's pointer
    }
};

I will call it like this:

Foo foo;
Bar bar;
bar.anotherFunc(foo.someFunc());

But how do I write anotherFunc()?

void anotherFunc(const Foo& foo)
{
    fooPointer = &foo;  // <-- This doesn't work as foo is still const
}

Upvotes: 1

Views: 1718

Answers (2)

Mike Seymour
Mike Seymour

Reputation: 254691

The best answer is "don't do that, it's crazy". Either change the pointer type to const Foo* if it won't be used to modify the object, or the reference to Foo& if it will.

If you genuinely have a good reason to discard const-correctness, then there's a cast for that

fooPointer = const_cast<Foo*>(&foo);

but first consider why you want to do this, and how you can prevent the error of ending up with a non-const pointer to a const object. Or worse, a pointer to a temporary that lingers after its demise.

Upvotes: 6

JBL
JBL

Reputation: 12907

You could accept it directly as a pointer or non-const reference to Foo.

The way you're passing your Foo, you won't be able to do that without a const_cast, which feels wrong except in a few precise case.

As you're storing it with a pointer to non-const Foo, you should take the parameter as non-const also. Think about the interface: your struct has a function that takes a Foo with const qualification, implying "I will only peak at it", but then your function casts it and stores it in a way that allows modification. It's a bit like lying to users of your code (and you are the first user of your code).

Upvotes: 4

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