Help porting c++ to php code

Question

I need if possible port this function to c++:

static unsigned char bux_code[3] = {0xFC, 0xCF, 0xAB};

void bux_convert(char* buf, int size)
{
        int n = 0;
        // ----
        for (n=0;n<size;n++)
        {
                buf[n]^=bux_code[n%3];
        }
}

What i have done:

$bux_code = array("\xFC", "\xCF", "\xAB");

function bux_convert($string)
{
    $size = strlen($string);

    for ($n = 0; $n < $size; $n++) {
        $string[$n] ^= $bux_code[$n % 3];
    }

    return $string;
}

var_dump ( bux_convert("£Ì•½Ï¼«ü") );

But i got this error: Cannot use assign-op operators with overloaded objects nor string

Hammerite · Accepted Answer · 2011-03-09 18:19:44Z

1

In PHP you can't always address an offset of a string like $mystring[5] or $mystring[$k] and then treat the resulting character as a numeric value in every respect. Here's a way you could get around it: turn the string into an array at the start of bux_convert(), using str_split(), and then turn it back into a string at the end, using implode().

Also, $bux_code is a global variable and won't be visible inside the function. Either move it inside the function or declare it as a global.

function bux_convert($string)
{
    $bux_code = array(0xFC, 0xCF, 0xAB);

    $size = strlen($string);
    $string = array_map('ord', str_split($string));

    for ($n = 0; $n < $size; $n++) {
        $string[$n] ^= $bux_code[$n % 3];
    }

    return implode('', array_map('chr', $string));
}

var_dump ( bux_convert("£Ì•½Ï¼«ü") );

edited Mar 9, 2011 at 18:19

answered Mar 9, 2011 at 18:09

Hammerite

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1 Comment

Matt Jr. Over a year ago

Wow! Thank you (and all others)!

madmik3 · Accepted Answer · 2011-03-09 18:03:17Z

0

try this.

 $bux_code = array(hexdec("FC"), hexdec("CF"), hexdec("AB"));

answered Mar 9, 2011 at 18:03

madmik3

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Comments

David Gelhar · Accepted Answer · 2011-03-09 18:05:09Z

0

The error is complaining about the "^=" assignment operator.

So, try doing what the error message suggests, and rewrite the assignment without using the assign-op operator:

$string[$n] = $string[$n] ^ $bux_code[$n % 3];

answered Mar 9, 2011 at 18:05

David Gelhar

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1 Comment

Matt Jr. Over a year ago

Ok, working now. But function returning "2212212212". In c++ it returning normal word

GolezTrol · Accepted Answer · 2011-03-09 18:08:02Z

0

You need to explicitly port globals to the function, so on the first line of your function put the line:

global $bux_code;

Now, the variable is not found, is undeclared, and therefor xor cannot operator on it.

answered Mar 9, 2011 at 18:08

GolezTrol

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