local var destroy when throw e for refer in c++? - c++

Question

Code sample one:
try {
exception e;
throw e;
} catch(exception& refer)
Code sample two:
exception& method()
{
exception e;
return e;
}
Some books mentioned that code sample one is ok, code two is wrong due to e in it is a local variable and will destroy when the function ends, but my question is why is sample code one okay? Isn't the e a local variable too?

Answer 1

Yes, in sample one, e is local to the try block, and is destroyed upon exiting that scope. But when you throw, the compiler makes a copy (or move), who's lifetime is extended to the end of the catch block, and it is this copy to which the reference refer refers.
For figuring out what the compiler does in cases like this, I always like to pull out my Noisy class.
#include <iostream>
class Noisy
{
public:
Noisy()
{ std::cout << "Noisy default construct\n"; }
Noisy(Noisy const&)
{ std::cout << "Noisy copy\n"; }
Noisy(Noisy&&)
{ std::cout << "Noisy move\n"; }
~Noisy()
{ std::cout << "Noisy destroy\n"; }
Noisy& operator=(Noisy const&)
{ std::cout << "Noisy copy assign\n"; return *this; }
Noisy& operator=(Noisy&&)
{ std::cout << "Noisy move assign\n"; return *this; }
void swap(Noisy&)
{ std::cout << "Noisy swap\n"; }
};
int main(int argc, char* argv[])
{
try
{
std::cout << "in try block\n";
Noisy n;
std::cout << "about to throw n\n";
throw n;
std::cout << "end of try block\n";
}
catch (Noisy & n)
{
std::cout << "in catch block\n";
}
std::cout << "after catch\n";
}