Printing an uninitialized reference gives 1. Why?

Question

This is kind of a follow-up question to my previous question Why is my exception still being thrown after being caught?

I have a reference data member r of class S which attempts to be initialized with the result of the function call f. f throws so I have set up a try-catch block. I don't expect r to be given a value as the exception is thrown before the function returns. But when I print out r in the catch block I get 1 as a result.

#include <iostream>
#include <stdexcept>

int& f() { throw std::invalid_argument("invalid"); static int i(50); return i; }

struct S
{
    S()
        try : r(f())
    { }
        catch(...)
    {
        std::cout << r;
    }
    int& r;
};

int main()
{
    try { S s; } catch(...) { }
}

Output:

I don't know why I am getting 1 as opposed to a random value or 0 like a regular variable. Is a result of 1 mandated by the Standard or is this just Undefined Behavior? I would really like to know. Thanks.

I would say it's undefined. Static locals are initialized when first run into, which never happens. Unless it's something strange like at the beginning of the function on its first call. — chris, Dec 03 '13 at 20:08
Aha, evidence: http://coliru.stacked-crooked.com/a/23b801f36fad4922. Feel free to add this to http://cppquiz.org :) — chris, Dec 03 '13 at 20:12

score 4 · Accepted Answer · answered Dec 03 '13 at 20:10

4

Since r is never initialized because the function that is going to initialize it throws an exception, you are printing an uninitialized reference, which will be undefined behavior.

answered Dec 03 '13 at 20:10

Zac Howland

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Printing an uninitialized reference gives 1. Why?

1 Answers1