CODEWITHSUNDEEP
c++pointerssmart-pointersimplicit-conversionlanguage-design
kzaiwo
kzaiwo

Reputation: 1748

Smart pointers cannot be used as raw pointers automatically?

I'm trying to convert codes to use smart pointers instead of raw pointers. But was wondering - is it really not possible for smart pointers to be used as raw pointers?

In this example:

#include <iostream>
#include <string>
#include <memory>

void printer(int* x)
{
    std::cout << "x = " << x << std::endl;
}

int main()
{   
    auto x = std::make_unique<int>(5);
    printer(x);
    return 0;
}

I get an error:

In function 'int main()': 14:14: error: cannot convert
'std::unique_ptr<int, std::default_delete<int> >' to 'int*' for argument '1'
to 'void printer(int*)'

How should this be coded? Thanks!

Upvotes: 2

Views: 887

Answers (2)

user3152801
user3152801

Reputation: 81

std::make_unique<>() returns a std::unique_ptr, which makes variable x of type std::unique_ptr<int>. These smart pointers are not just raw pointers and you cannot use them just like raw pointers. You should instead use the get() method on x to get the int* it is pointing to.

Upvotes: 0

StoryTeller - Unslander Monica
StoryTeller - Unslander Monica

Reputation: 170074

There cannot be an implicit conversion because an implicit conversion doesn't have context to do the right thing. printer takes a non-owning pointer, but what about other functions that need to take ownership? For instance

void kill(int *p) { delete p; }

Silently doing the conversion when calling kill(p); would be the wrong thing! It will lead to a double delete. The pointer has no way to know it should stop owning the object, it has to be told via a call to release. Anything else would be wrong.

Since smart pointers are all about enforcing explicit ownership semantics, the conversion to a raw pointer needs to be explicit too, and have a name that conveys how the ownership is affected.

Upvotes: 2

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