将javascript代码转换为PHP：确定返回（1）在javascript中的含义

Question

I have this code in JS

function calc(bth){
    var y,cont;
    y=(Math.abs(1965-bth))/4;
    y=y-Math.floor(y);    
    if(y){
      cont=calc(bth+1);
      return(cont+1);
    } else {
      return(1); // <-- what is the meaning of this line?
    }
 }

I converted it to PHP as

function calc($yr)
{       
    $y = (abs(1965 - $yr))/4;
    $y = $y- floor($y);    
    if($y)
    {
       $cont= calc($yr+1);       
       return ($cont + 1);
    }
    else
    {
        return (1); //<--- stuck here
    }
}

The Problem

I realized that return (1) in javascript is not the same as return (1) in PHP. I went over some SO topics regarding this topic but didn't find a suitable match for my kind of scenario.

In my context, what would return (1) return in javascript? I would also appreciate if the values of return (0) and return (-1) in javascript are made known to me so that I get acquainted with this information.

Answer 1

Don't overcomplicate the problem. It will simply return 1 , as if you write return 1 .

Most probably brackets were left from copying of return(cont+1) .

Answer 2

return is not a function, () is not necessary for return statement.

Just do with return cont + 1; and return 1;