Exception safety and make_unique

Question

Just to clarify, using make_unique only adds exception safety when you have multiple allocations in an expression, not just one, correct? For example

void f(T*);

f(new T);

is perfectly exception safe (as far as allocations and stuff), while

void f(T*, T*);

f(new T, new T);

is not, correct?

Answer 1

Not only when you have multiple allocations, but whenever you can throw at different places. Consider this:

f(make_unique<T>(), function_that_can_throw());

Versus:

f(unique_ptr<T>(new T), function_that_can_throw());

In the second case, the compiler is allowed to call (in order):

• new T
• function_that_can_throw()
• unique_ptr<T>(...)

Obviously if function_that_can_throw actually throws then you leak. make_unique prevents this case.

And of course, a second allocation (as in your question) is just a special case of function_that_can_throw().

As a general rule of thumb, just use make_unique so that your code is consistent. It is always correct (read: exception-safe) when you need a unique_ptr, and it doesn't have any impact on performance, so there is no reason not to use it (while actually not using it introduces a lot of gotchas).

Answer 2

As of C++17, the exception safety issue is fixed by a rewording of [expr.call]

The initialization of a parameter, including every associated value computation and side effect, is indeterminately sequenced with respect to that of any other parameter.

Here indeterminately sequenced means that one is sequenced before another, but it is not specified which.

f(unique_ptr<T>(new T), function_that_can_throw());

Can have only two possible order of execution

1. new T unique_ptr<T>::unique_ptr function_that_can_throw
2. function_that_can_throw new T unique_ptr<T>::unique_ptr

Which means it is now exception safe.