pass smart pointers as a pointer of pointer to a function

Question

I'm trying to pass a smart pointer as a parameter of a function which takes a pointer of a pointer. I was wondering if there is any proper solution for that.

foo(Class** input)
{
   // Do something
}

myClass = std::make_Unique<Class>();

foo(&myClass.get())

Why would you want to do that? Manipulating the pointer that a smart pointer manages sounds a bit dangerous. — lubgr
– lubgr, Commented Sep 12, 2018 at 10:38

MSalters · Accepted Answer · 2018-09-12 10:40:51Z

2

That won't work, and it's a good thing that it doesn't.

There are roughly two reasons why you need to pass a Foo**. It could be a function that wants a 2D array (array of pointers to arrays), or the function has a Foo*as output (should have used a Foo*& then). In your case, it appears that you're dealing with the array case, since the argument is called input.

answered Sep 12, 2018 at 10:40

MSalters

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8 Comments

luk32 Over a year ago

I'd be reluctant to call T** a 2D array. Array of pointers has a different memory layout than a 2D array, i.e. T[N][M]. The difference is quite substantial.

Mohsen67 Over a year ago

Thanks for your response. It's actually a return value of the function foo. but I can't change the signature.

n. m. could be an AI Over a year ago

@Mohsen67 Why would you want to pass a pointer to a function that wants to return a pointer?

Mohsen67 Over a year ago

What else can I do? either pass a raw pointer or I thought it's better to use smart pointer

MSalters Over a year ago

@Mohsen67: Well , you are right, but the signature isn't right.

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