将简单的函数从jS转换为PHP

Question

I have a simple function hash function in JS that I'm struggeling with converting into PHP (for server side validation), it is a simple math function, and I'm not a PHP expert, could any of you shed some light into why I can't have ($variable)(function) in PHP next to each other?

Here is the function in jS:

function claculate(a,b){
    var g=function(a){
      return Math.round(a/Math.pow(10,Math.round(Math.LOG10E*Math.log(a)))*10)
    };
    var f=function(a){
        var h=function(b){
            return b(a)
        };
        var j=function(b){
            return g(a)
        }; 
        return a<b?j:h
     };
     var o=((f(a)(g)%2)||(f(a*b)(g)%7*1/7)||11)^f(b)(g);
     return another_function(a,o);  
}

And here is my version in PHP, but there is an error on the $o line, and I'm not sure why ...

function claculate($a, $b){
   $g = function ($r){
       return round($r / pow(10, round(M_LOG10E * log($r))) * 10);
   };
   $f = function ($a) use (&$a, &$b, &$g) {
       $h = function ($b) use (&$b, &$a) {
           return $b($a);
        };
        $j = function ($b) use (&$b, &$g, &$a) {
           return $g($a);
        };
        return ($a < $b) ? $j : $h;
    };
    // $o throws error
    $o = $f($a)($g) % 2 || $f($a * $b)($g) % 7 * 1 / 7 || 11 ^ $f($b)($g);
    return another_function($a, $o);
}

Any thoughts or right direction is more than welcomed! Thanks

Answer 1

In general, this works the same:

function g($r)
{
  return round($r / pow(10, round(M_LOG10E * log($r))) * 10);
}

function claculate($a, $b)
{
  $o = ((g($a) % 2) || (g($a * $b) % 7 * 1 / 7) || 11) ^ g($b);
}

The first problem is in difference when evaluating OR statement. (JS returns number, PHP bool.) But this can be solved by using of some conditions.

The worst is that JS uses 2 more digits for real numbers.