Static initialization of non-POD that throws

[Old Wolf]

In the following code:
1) Is the exception allowed to be thrown even if the
user never enters '1'?
2) If the user enters '1' a second time, what happens?

How does the compiler typically implement all this?

#include <iostream>

int x = 0;

struct S
{
void foo() { std::cout << "In foo" << std::endl; }
S() { if (x++ == 0) throw 1; }
};

int main()
{
while (1) switch( std::cin.get() )
{
case '1': try { static S s; s.foo(); } catch(...) {} break;
case EOF: return 0;
}
}

[anon]

Old Wolf wrote:

In the following code:
1) Is the exception allowed to be thrown even if the
user never enters '1'?

Constructors can throw exceptions, but in this case, the construction
will first wait for cin to finish

2) If the user enters '1' a second time, what happens?

Since i is not 0 anymore, nothing will be thrown from the constructor
and the object will be constructed.
Third and fourth and further 1s will not call constructor, because you
declared your object static.

>
How does the compiler typically implement all this?

#include <iostream>

int x = 0;

struct S
{
void foo() { std::cout << "In foo" << std::endl; }
S() { if (x++ == 0) throw 1; }
};

int main()
{
while (1) switch( std::cin.get() )
{
case '1': try { static S s; s.foo(); } catch(...) {} break;
case EOF: return 0;
}
}