javascript - JS why is other code ignored and only return executed -

i got code gave same result:

var number = 0; function a(){     number++; //outside return function     return function(){         return number;     }; } var b = a(); //outside loop for(i=0;i<10;i++){     console.log("loop: "+(i+1)+" :"+b()); } 

i found solution (1):

var number = 0; function a(){     number++; //outside return function     return function(){         return number;     }; } for(i=0;i<10;i++){     var b = a(); //inside loop     console.log("loop: "+(i+1)+" :"+b()); } 

and apparently should've gotten "correct" solution (2):

var number = 0; function a(){     return function(){         number++; //inside return function         return number;     }; } var b = a(); //outside loop for(i=0;i<10;i++){     console.log("loop: "+(i+1)+" :"+b()); } 

in solution 1: thought putting var b in for-loop, "refresh" var number every loop, , did because every loop 1 added.

in solution 2: don't see why number++ should added return function looped. i.m.o. whole a() function run through in original question , not code in return method? not, why?

in first version a() called once, number++ called once. in second version call a() every iteration number++ called on every iteration. in first version b same function , in second new 1 every time call a().