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+--- Forum: AUTOSAR C++:2014 rules (https://forum.misra.org.uk/forumdisplay.php?fid=185)
+--- Thread: A18-9-4 (/showthread.php?tid=1714)
A18-9-4 - cgpzs - 15-10-2024
Hi!
I wonder if this code example technically violates A18-9-4?
Code:
#include <memory>
struct Foo
{
explicit Foo(int);
};
int main()
{
int val{};
auto x = std::make_unique<Foo>(val);
++val; // non-compliant?
}Since internally, std::make_unique takes the input arguments as "Args&&..." and std::forward's them into the constructor of the class.
Thanks!
RE: A18-9-4 - misra cpp - 22-10-2024
The short (and unhelpful) answer is that the rule only talks about uses of std::foward, and this doesn't appear in your example, so as far as the example is concerned, the rule doesn't apply. Its a general principle that rules don't consider how library functions may be implemented.
We believe that the intent was to limit use after std::foward 'within the same function body, not in general'. So, even if there is a std::forward inside std::make_unique, your example is still OK (apart from the fact it doesn't compile - Foo needs a copy constructor).
The following example may clarify further:
#include <utility>
template <typename T> void foo (T && arg)
{
auto j = std::forward<T> (arg);
// non-compliant to use 'arg' here
}
void bar ()
{
int i = 0;
foo (i);
++i; // Compliant
foo (std::move(i));
++i; // Rule A18-9-4 doesn't apply, but
// non-compliant with A12-8-3 because
// of read of moved from object
}
RE: A18-9-4 - cgpzs - 23-10-2024
Understood, thank you for the response! It would be great to add a similar clarification to the equivalent rule 28.6.3 in MISRA C++:2023.