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This function is from my professor's notes:

int ints_is_sorted_r(int* a, int n){
    return n <= 1 || (a[0] <= a[1] && ints_is_sorted_r(a+1, n-1));
}

This is my version with many printfs to see what it does

int ints_is_sorted_r(int* a, int n){

    printf("n <= 1 = %d\n",(n <=1));
    printf("a[0] <= a[1] = %d <= %d\n",a[0],a[1]);
    ints_println_special(a,n);
    return n <= 1 || (a[0] <= a[1] && ints_is_sorted_r(a+1, n-1));

}

(ints_println_special() is a function from my prof's library that prints the whole array)

My question is, how does the recursion stop? Here's the output from my test:

enter image description here

My guess is that when you have an OR condition and the first one is true it doesn't waste time looking for the right side of the condition because it knows it's going to be true anyway. Am I correct?

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SilenceOnTheWire
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  • Google for "short circuit evaluation". See https://softwareengineering.stackexchange.com/questions/201896/what-is-short-circuiting-in-c-like-languages. – Jonathon Reinhart Dec 08 '19 at 21:56
  • @Carcigenicate: OR if first evaluand is TRUE // AND if first evaluand is FALSE will ***both*** short-circuit. – Adrian Mole Dec 08 '19 at 22:04
  • @Carcigenicate No worries! If there were "Whoops" badges on S.O., I'd have earned several gold ones, already. Also, your comment helped me improve my answer (the PS) - hope you don't mind. – Adrian Mole Dec 08 '19 at 22:16
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    Tip: posting text as a picture rather than text attracts down votes. Text as text is more useful. – chux - Reinstate Monica Dec 08 '19 at 22:52

1 Answers1

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My guess is that when you have an OR condition and the first one is true it doesn't waste time looking for the right side of the condition because it knows it's going to be true anyway. Am I correct?

Yes, you are correct! It's called "short-circuiting" and there are several good posts here on Stack Overflow that explain and discuss it, like this one.

PS: Note that there could also be (probably will be) such short-circuiting, internally, in the part after the || operator; so, in (a[0] <= a[1] && ints_is_sorted_r(a+1, n-1)), if the first comparison, a[0] <= a[1], is FALSE, then the second won't be evaluated, and the function won't be called recursively in that case.

Adrian Mole
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  • If you think an answer on another question would explain it better, you should vote to close as duplicate of that question instead of answering. – S.S. Anne Dec 08 '19 at 22:35