Value of int i = i ^ i ; Is it always zero or undefined behavior?

Question

in following program, is output always zero, or undefined behavior?

#include<iostream>

int main()
{
    int i= i ^ i ;
    std::cout << "i = " << i << std::endl;
}

with gcc 4.8.0 this code success compiled, and output is 0.

Answer 1

int i= i ^ i ;

Since i is an automatic variable (i.e it is declared in automatic storage duration), it is not (statically) initialized yet you're reading its value to initialize it (dynamically). So your code invokes undefined behaviour.

Had you declared i at namespace level or as static, then your code would be fine:

• Namespace level

  int i = i ^ i; //declared at namespace level (static storage duration)
  
  int main() {}
  

• Or define locally but as static:

  int main()
  {
       static int i = i ^ i; //static storage duration
  }
  

Both of these code are fine, since i is statically initialized, as it is declared in static storage duration.

Answer 2

Undefined behavior. Uninitialized garbage doesn't actually have to be an unknown but valid value of the given type. On some architectures (specifically Itanium), uninitialized garbage can actually cause a crash when you try to do anything with it. See http://blogs.msdn.com/b/oldnewthing/archive/2004/01/19/60162.aspx for an explanation of how IA64's Not a Thing can mess you up.