javascript - Understanding variable mistakes within closures? -

i learning closures. example given common mistake made when making closure:

function assigntorpedo(name, passengerarray) {     var torpedoassignment;     (var = 0; i<passengerarray.length; i++) {         if (passengerarray[i] == name) {             torpedoassignment = function() {                 alert("ahoy, " + name + "!\n" +                 "man post @ torpedo #" + (i+1) + "!");             };         }     }     return torpedoassignment; } 

since loop completes before closure returned, value not match name. so, understand loop continues on before return happens.

my question comes this, example of correct way things:

function maketorpedoassigner(passengerarray) {     return function (name) {         (var = 0; i<passengerarray.length; i++) {             if (passengerarray[i] == name) {                 alert("ahoy, " + name + "!\n" +                 "man post @ torpedo #" + (i+1) + "!");             }         }     }; } 

i don't understand why in above example loop wouldn't continue past first time finds match, result in mismatched i. understand return stops function, don't understand connection between return , first match since don't happen (visually). understand how code knew stop if return within if function or loop.

i don't understand why in above example loop wouldn't continue past first time finds match

it would.

which result in mismatched i.

it wouldn’t, because checks if (passengerarray[i] == name) every time. that’s wasteful, though; it’s unusual fix. better way pass index:

function maketorpedoassigner(passengerarray, i) {     return function (name) {         alert("ahoy, " + name + "!\n" +               "man post @ torpedo #" + (i+1) + "!");     }; }  function assigntorpedo(name, passengerarray) {     (var = 0; i<passengerarray.length; i++) {         if (passengerarray[i] == name) {             return maketorpedoassigner(passengerarray, i);         }     } }