Should a theoretically, but not practically, throwing function be declared noexcept?

Question

Is it safe to declare the following function noexcept even though v.at(idx) could theoretically throw a out_of_range exception, but practically not due to the bounds check?

int get_value_or_default(const std::vector<int>& v, size_t idx) noexcept {     if (idx >= v.size()) {         return -1;     }     return v.at(idx); } 

Answer 1

What is you definition of "safe"? If you throw an exception in a function marked asnoexcept or noexcept(true) then your program will be terminated (standard 15.4.9)

Whenever an exception is thrown and the search for a handler (15.3) encounters the outermost block of a function with an exception-specification that does not allow the exception, then,

• if the exception-specification is a dynamic-exception-specification, the function std::unexpected() is called (15.5.2),
• otherwise, the function std::terminate() is called (15.5.1).

As long as that is acceptable then you are fine. If you can not tolerate the program terminating then it is not safe.

Answer 2

It is safe. Declaring a function noexcept would result in immediate program termination (by a call to terminate()) if an exception occurs. As long as no exception is thrown everything is fine.