将 JavaScript function 转换为 PHP

Question

. Answers to this question are eligible for a +100 reputation bounty. szmegma wants to to this question:

Hopefully, someone can provide a working version of PHP.

I have a JS function, with array inputs.

For example:

x=[ 239709880, 250229420, 109667654, 196414465, 13098 ]
y=[ 78135241, 54642792, 249 ]

OR:

x=[ 0, 0, 0, 0, 0, 0, 1 ]
y=[ 78135241, 54642792, 249 ]

OR:

x=[ 49 ]
y=[ 33 ]

function bdiv(x,y) {
    var n=x.length-1, t=y.length-1, nmt=n-t, arr = []
    if(n < t || n==t && (x[n]<y[n] || n>0 && x[n]==y[n] && x[n-1]<y[n-1])) {
        arr['q']=[0]
        arr['mod']=x
        return arr
    }
    if(n==t && toppart(x,t,2)/toppart(y,t,2) <4) {
        var q=0, xx
        for(;;) {
            xx=bsub(x,y)
            if(xx.length==0) break
            x=xx; q++
        }
        arr['q']=[q]
        arr['mod']=x
        return arr
    }
    var shift, shift2
    shift2=Math.floor(Math.log(y[t])/log2)+1
    shift=bs-shift2

    if(shift) {
        x=x.concat()
        y=y.concat()
        for(i=t; i>0; i--) y[i]=((y[i]<<shift) & bm) | (y[i-1] >> shift2); y[0]=(y[0]<<shift) & bm
        if(x[n] & ((bm <<shift2) & bm)) { x[++n]=0; nmt++; }
        for(i=n; i>0; i--) x[i]=((x[i]<<shift) & bm) | (x[i-1] >> shift2); x[0]=(x[0]<<shift) & bm
    }
    var i, j, x2, y2,q=zeros(nmt+1)

    y2=zeros(nmt).concat(y)

    for(;;) {
        x2=bsub(x,y2)
        if(x2.length==0) break
        q[nmt]++
        x=x2
    }
    var yt=y[t], top=toppart(y,t,2)
    for(i=n; i>t; i--) {
        m=i-t-1
        if(i >= x.length)
            q[m]=1
        else if(x[i] == yt)
            q[m]=bm
        else
            q[m]=Math.floor(toppart(x,i,2)/yt)
        topx=toppart(x,i,3)
        while(q[m] * top > topx)
            q[m]--
        y2=y2.slice(1)
        x2=bsub(x,bmul([q[m]],y2))
        if(x2.length==0) {
            q[m]--
            x2=bsub(x,bmul([q[m]],y2))
        }
        x=x2
    }
    if(shift){
        for(i=0; i<x.length-1; i++)
            x[i]=(x[i]>>shift) | ((x[i+1] << shift2) & bm);
        x[x.length-1]>>=shift
    }
    while(q.length > 1 && q[q.length-1]==0)
        q=q.slice(0,q.length-1)
    while(x.length > 1 && x[x.length-1]==0)
        x=x.slice(0,x.length-1)
    arr['q']=q
    arr['mod']=x
    return arr;
}

What I have done under 5 days so far in PHP:

function bdiv($x,$y){
    global $bs, $bm, $bx2, $bx, $bd, $bdm, $log2;
    $arr=[];
    $n=count($x)-1;
    $t=count($y)-1;
    $nmt=$n-$t;

    if($n < $t || $n==$t && ($x[$n]<$y[$n] || $n>0 && $x[$n]==$y[$n] && $x[$n-1]<$y[$n-1]))
        return ['q'=>[0], 'mod'=>$x];

    if($n==$t && toppart($x,$t,2)/toppart($y,$t,2) <4){
        $q=0;
        for(;;){
            $xx=bsub($x,$y);
            if(count($xx)==0)
                break;
            $x=$xx;
            $q++;
        }
        return ['q'=>[$q], 'mod'=>$x];
    }

    $shift2=floor(log($y[$t])/$log2)+1;
    $shift=$bs-$shift2;
    if($shift){

/////////////////////////////////////////////////// the problem start here with the concat()
        array_push($x,$x);
        array_push($y,$y);

        for($i=$t; $i>0; $i--)
            $y[$i]=(($y[$i] << $shift) & $bm) | ($y[$i-1] >> $shift2);
        $y[0]=($y[0] << $shift) & $bm;
        if($x[$n] & (($bm << $shift2) & $bm)){
            $x[++$n]=0;
            $nmt++;
        }
        for($i=$n; $i > 0; $i--)
            $x[$i]=(($x[$i] << $shift) & $bm) | ($x[$i-1] >> $shift2);
        $x[0]=($x[0] << $shift) & $bm;
    }
    $q=zeros($nmt+1);

    array_push($arr, zeros($nmt));
    array_push($arr, $y);
    $y2=array_merge(...$arr);

    for(;;){
        $x2=bsub($x,$y2);
        if(count($x2)==0)
            break;
        $q[$nmt]++;
        $x=$x2;
    }

    $yt=$y[$t];
    $top=toppart($y,$t,2);

    for($i=$n; $i>$t; $i--){
        $m=$i-$t-1;
        if($i >= count($x))
            $q[$m]=1;
        else if($x[$i] == $yt)
            $q[$m]=$bm;
        else
            $q[$m]=floor(toppart($x,$i,2)/$yt);

        $topx=toppart($x,$i,3);
        while($q[$m] * $top > $topx)
            $q[$m]--;

        $y2=array_slice($y2,1);
        $x2=bsub($x,bmul([$q[$m]],$y2));

        if(count($x2)==0){
            $q[$m]--;
            $x2=bsub($x,bmul([$q[$m]],$y2));
        }
        $x=$x2;
    }

    if($shift){
        for($i=0; $i<count($x)-1; $i++)
            $x[$i]=($x[$i] >> $shift) | (($x[$i+1] << $shift2) & $bm);
        $x[count($x)-1] >>= $shift;
    }

    while(count($q) > 1 && $q[count($q)-1]==0)
        $q=array_slice($q, 0, count($q)-1);
    while(count($x) > 1 && $x[count($x)-1]==0)
        $x=array_slice($x, 0, count($x)-1);

    return ['q'=>$q, 'mod'=>$x];
}

So as I marked in the PHP code I have a problem with the array_push($x,$x) , seems like this is not the equivalent of x=x.concat() . Array_push add the whole current $x values as a new element to the existing $x array:

$x=[ 1, 2, 3 ];
array_push($x,$x);
then $x will be [ 1, 2, 3, [ 1, 2, 3 ] ]

If I try to flatten the array ( $x=array_merge(...$x); ) then a new PHP error shows up: array_merge(): Argument #1 is not an array

I would really appreciate it if anyone have any idea, how to convert properly this JS function to a PHP version. Thanks in advance.

Answer 1

replace array_push with array_merge.

This will return the merged array then store the result in $x;

array_merge is meant for arrays. it will take the values from one array and append it to the other. just like concat does in JS.

... splits the array ($x) into several values, this is not the correct input. it is the equivalent of array_merge(1,2,3) (ie arrays are not being inputted)

$x = array(1,2,3);
$x = array_merge($x,$x);
var_dump($x); //output: [1,2,3,1,2,3]

echo $x[4]; // output: 2

if i've misunderstood the question please let me know.

Answer 2

Use array_merge instead of array_push ie

  $x = array(1, 2, 3);
       print_r(array_merge($x,$x));

Answer 3

I am confused why your JavaScript code even has:

x=x.concat()
y=y.concat()

They don't serve any purpose other than assigning a new copy of same array to itself. If it was intended to avoid modification on the original array, then you can simply replace those two lines of code with:

$x = array_merge(array(), $x);
$y = array_merge(array(), $y);

They both serve exact same purpose.

At this point, I am not sure about rest of the php code that you wrote so far, but if it helps you in anyway, that is fine.

Answer 4

If what you are doing with x = x.concat() is attempting to ensure that the original passed array is not being modified by your function, you do not have to do anything to ensure that in your PHP version of this function because by default arguments will be copied rather than passed by reference . To force an array to be passed by reference in PHP, you must preceded the argument name with an ampersand ( & ). This can be demonstrated with the following program where we have defined two functions that each modify the first element of the passed array. In the first function, test1 , the array argument is copied so the original passed array remains unmodified. But in the second function, test2 , the array argument is passed by reference and when the function returns, the original passed array will have been modified. The only difference between the two functions is that in test1 the arguments is defined as $x and in test2 as &$x :

<?php

// $x is passed by value:
function test1($x)
{
    $x[0] = 9;
    print_r($x);
}

// $x is passed by reference:
function test2(&$x)
{
    $x[0] = 9;
    print_r($x);
}

$my_array = [0, 1, 2];
echo "Pass by value:\n";
test1($my_array); //$my_array remains unmodified
print_r($my_array);

echo "\n\nPass by reference:\n";
test2($my_array); //$my_array is modified
print_r($my_array);

Prints:

Pass by value:
Array
(
    [0] => 9
    [1] => 1
    [2] => 2
)
Array
(
    [0] => 0
    [1] => 1
    [2] => 2
)


Pass by reference:
Array
(
    [0] => 9
    [1] => 1
    [2] => 2
)
Array
(
    [0] => 9
    [1] => 1
    [2] => 2
)