Alexis Wilke Alexis Wilke - 1 year ago 103
C++ Question

Can a unique_ptr<>() initialization fail?

From the documentation of std::unique_ptr<>(), what may happen when initializing the pointer is not clear to me.

When allocating an

, it allocates a memory buffer to handle the reference counter. So I may get an

Could something similar happen when initializing a unique pointer?

I am asking the question because if it does, I may actually lose what I was attempting to have delete through the unique pointer. For example:

void deleter(FILE * f)

void func()
FILE * f(fopen("/tmp/random", O_CREAT | ...));
if(f == nullptr) ...handle error...
std::unique_ptr<FILE, decltype(&deleter)> raii_file(f, deleter);

So, if the initialization of
can throw, I may end up keeping the file
open forever. (I use
as an example, any similar resource could be affected.)

Opposed to this answer, I obviously cannot use
since I'm not just allocating memory.

Would it be safer to initialize the
before the
, and then save the value in there after?

std::unique_ptr<FILE, decltype(&deleter)> raii_file(nullptr, deleter);
FILE * f(fopen("/tmp/random", O_CREAT | ...));
if(f == nullptr) ...handle error...
raii_file = f;

Or would that have similar problems?

Answer Source

All of unique_ptr's constructors are noexcept. So no, there's no way that it can fail. If your Deleter type throws on copy/move, then the noexcept will catch it and call std::terminate.

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