How to handle an exception thrown from a function defined as noexcept?

Question

It is said that defining a function as noexcept will let the compiler do some optimizations to boost the program and if the function needs to throw the compiler will suppress that optimization.

Here's my simple snippet:

void func() noexcept(true){
    std::vector<int> vec{2, 4, 6, 8}; // vec is a vector of 4 ints
    std::cout << vec.at(5); // throws
}

int main(){

    try{
        func();
    }
    catch(std::exception const& e){
        std::cout << e.what() << '\n';
    }
    catch(...){
        //
    }

}

• Why I cannot catch the exception? I am sure because the noexcept exception specification. So what happens if a function is marked as noexcept but throws and how to catch its exception rather than letting the program call std::terminate() to end the program?

Answer 1

Why I cannot catch the exception?

Because the function is noexcept. This is exactly what it means. Declaring a function noexcept is a promise that nothing will ever be thrown out of the function. As a consequence, the compiler doesn't need to generate exception handling code. If that promise is broken, then the exception cannot be handled and only reasonable way to proceed is to terminate.

So what happens if a function is marked as noexcept but throws

If an exception is thrown out of the noexcept function, then the process is terminated.

and how to catch its exception rather than letting the program call std::terminate() to end the program?

There is no way to catch an exception thrown out of a noexcept function. Your options are:

• Accept that the process will be terminated if that happens.
• Don't let exceptions be thrown from noexcept functions, thus avoiding the termination. Note that throwing inside is fine, as long as nothing is thrown out.
• Don't declare functions that throw as noexcept, allowing you to handle the exception.