big o - Computing Time T(n) and Big-O with an infinite loop

Question

I'm confused on how to create a function T(n) to measure computing time for a nested infinite loop. Here is the code:

x=1;
for(int i = 0;i<n-1;i++){
     for(int j = 1; j<=x; j++){
        cout << j << endl;
        x*=2;
     }
}

So the inner loop will go on forever, and I am stuck trying create the function to represent its computing time. I have written that its computing time is T(n) = [Summation i=0 until (n-2)] (2^j). 2^j represents the value of x with the current value of j from the inner loop. After discussing this with my peers, we definitely agree that the computing time is certainly not dependent on the value of n. We also could be completely over-thinking this as the loop is infinite and there is simply no way to express its computing time. Any help is greatly appreciated.

Answer 1

Algorithm complexity is only defined for algorithms, which by (the most often accepted) definition must terminate. This process doesn't terminate (except "in practice" as Marcelo notes; i.e. as a program on a real machine vs. in theory on a theoretical Turing machine with an infinite tape and all the time in the world) so is not an algorithm. So it has no "algorithmic time complexity".

Trying to determine the algorithmic complexity for a non-algorithm is a futile exercise, as is trying to express its running time as a polynomial if it's an infinite process.