Pass a container of smart pointer as argument in C++

Question

I have a function like below

void functionA(unordered_map<string, classA*>* arg1);

I need to pass in

unordered_map<string, shared_ptr<classA>>

How could I pass in the container with shared_ptr to the function which takes in a container of raw pointer? I am using C++0x here.

Thanks

Answer 1

Typewise unordered_map<string, shared_ptr<classA>> and unordered_map<string, classA*> are unrelated types.

So you can't directly pass one where the other is expected.

You can:

• change the signature of your function, or

• copy the data to an object of the expected type.

By the way, instead of a pointer argument, consider a reference argument.

Also, consider adding const wherever possible – it generally helps making the code easier to understand, because you then have guarantee of non-modification.

Cheers & hth.,

Answer 2

您将不得不修改functionA，或构建一个其想要接收的类型的临时unordered_map。

Answer 3

Since you asked, here's what a template might look like for this function:

template <typename MAP_TYPE>
void functionA(const MAP_TYPE& arg1)
{
    // Use arg1 here
}

You could use a pointer and/or make it non-const if you must. As long as the map passed in uses a compatible key type and some kind of pointer to the right type as a value, it should compile and run. Whether the pointer is a native pointer or a smart pointer will likely not make a difference, depending on how you used it in your function implementation.