Dereferencing a shared pointer and assigning to it

Is it ok to dereference a shared pointer, assign and assign a new object to it like so:

void foo()
{
std::shared_ptr<std::string> x =
std::make_shared<std::string>();

bar(*x); // is this fine?
// x == bsl::string("WHATEVER")
}

void bar(string& y)
{
y = string("whatever");
}


You are making changes to the object managed by the pointer ( you are not making it manage a new object)
– M.M
Jun 29 at 2:47

Yes, so the above code is safe then? I don't end up modifying a temporary or something like that?
– Martin F
Jun 29 at 2:51

Yes this is all fine
– M.M
Jun 29 at 3:05

As an aside, consider using auto more. Repetition is error-prone.
– Deduplicator
Jun 29 at 8:39

auto

@MartinF "modifying a temporary" What about it? Would that be bad?
– curiousguy
2 days ago

2 Answers
2

Yes, this is valid. Operator * returns the result of dereferencing the stored (raw) pointer.

*

Dereferencing a (raw) pointer does not make a copy or return a temporary: dereferencing a pointer when passing by reference

It is fine to do it but I would use the shared_ptr::get() operator in order to be more obvious regarding the fact that we are not using a raw pointer.

shared_ptr::get()

void foo()
{
std::shared_ptr<std::string> sharedPtr = std::make_shared<std::string>();
bar(*sharedPtr.get());
}


Why would you want to go out of your way to call out that you are using a smart-pointer?
– Deduplicator
Jun 29 at 8:38

I am not going out of my way, I am just using the default operator when returning the stored pointer. Anyway, it is more a question of taste for me
– marti_dms
Jun 29 at 8:47

You are using an extraneous .get(). That is going out of your way.
– Deduplicator
Jun 29 at 9:27

.get()

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