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How can I make the PHP version of Bitwise XOR match the javascript version?

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0

How can I make the PHP version of Bitwise XOR match the javascript version?

The javascript version gives me 1080

(f6o5y5^t0u1)+(p6t0k1^l2v2)+(j0u1f6^d4d4)+(p6t0k1^l2v2)

While the PHP gives me 0000

$a = f6o5y5^t0u1 & 0xffffffff;
if ($a & 0x80000000)
 $a -= 0x100000000;

$b = p6t0k1^l2v2 & 0xffffffff;
if ($b & 0x80000000)
 $b -= 0x100000000;

$c = j0u1f6^d4d4 & 0xffffffff;
if ($c & 0x80000000)
 $c -= 0x100000000;

$d = p6t0k1^l2v2 & 0xffffffff;
if ($d & 0x80000000)
 $d -= 0x100000000;

echo $a.$b.$c.$d; 

javascript php bit-manipulation xor bitwise-xor

share|improve this question

asked Mar 23 at 8:56

William CrossWilliam Cross

333

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What are the values of all input variables?
  
  – Gnudiff
  Mar 23 at 8:59
  

  
  
  
  

add a comment | 

0

How can I make the PHP version of Bitwise XOR match the javascript version?

The javascript version gives me 1080

(f6o5y5^t0u1)+(p6t0k1^l2v2)+(j0u1f6^d4d4)+(p6t0k1^l2v2)

While the PHP gives me 0000

$a = f6o5y5^t0u1 & 0xffffffff;
if ($a & 0x80000000)
 $a -= 0x100000000;

$b = p6t0k1^l2v2 & 0xffffffff;
if ($b & 0x80000000)
 $b -= 0x100000000;

$c = j0u1f6^d4d4 & 0xffffffff;
if ($c & 0x80000000)
 $c -= 0x100000000;

$d = p6t0k1^l2v2 & 0xffffffff;
if ($d & 0x80000000)
 $d -= 0x100000000;

echo $a.$b.$c.$d; 

javascript php bit-manipulation xor bitwise-xor

share|improve this question

asked Mar 23 at 8:56

William CrossWilliam Cross

333

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What are the values of all input variables?
  
  – Gnudiff
  Mar 23 at 8:59
  

  
  
  
  

add a comment | 

How can I make the PHP version of Bitwise XOR match the javascript version?

The javascript version gives me 1080

(f6o5y5^t0u1)+(p6t0k1^l2v2)+(j0u1f6^d4d4)+(p6t0k1^l2v2)

While the PHP gives me 0000

$a = f6o5y5^t0u1 & 0xffffffff;
if ($a & 0x80000000)
 $a -= 0x100000000;

$b = p6t0k1^l2v2 & 0xffffffff;
if ($b & 0x80000000)
 $b -= 0x100000000;

$c = j0u1f6^d4d4 & 0xffffffff;
if ($c & 0x80000000)
 $c -= 0x100000000;

$d = p6t0k1^l2v2 & 0xffffffff;
if ($d & 0x80000000)
 $d -= 0x100000000;

echo $a.$b.$c.$d; 

javascript php bit-manipulation xor bitwise-xor

share|improve this question

asked Mar 23 at 8:56

William CrossWilliam Cross

333

(f6o5y5^t0u1)+(p6t0k1^l2v2)+(j0u1f6^d4d4)+(p6t0k1^l2v2)

$a = f6o5y5^t0u1 & 0xffffffff;
if ($a & 0x80000000)
 $a -= 0x100000000;

$b = p6t0k1^l2v2 & 0xffffffff;
if ($b & 0x80000000)
 $b -= 0x100000000;

$c = j0u1f6^d4d4 & 0xffffffff;
if ($c & 0x80000000)
 $c -= 0x100000000;

$d = p6t0k1^l2v2 & 0xffffffff;
if ($d & 0x80000000)
 $d -= 0x100000000;

echo $a.$b.$c.$d; 

share|improve this question

asked Mar 23 at 8:56

William CrossWilliam Cross

333

share|improve this question

asked Mar 23 at 8:56

William CrossWilliam Cross

333

asked Mar 23 at 8:56

William CrossWilliam Cross

333

asked Mar 23 at 8:56

William CrossWilliam Cross

333

1 Answer
1

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oldest

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0

The syntax is the same, the implementation in your sample is different. For example, the following Javascript:

var x = (1^2)+(3^4)+(5^6)+(4^8);
console.log(x);

Will output x as 25;

The PHP version is the same. For example:

<?=(1^2)+(3^4)+(5^6)+(4^8)?>

Will also render 25.

My guess is you do not have the same values or your syntax is the different in the different languages.

share|improve this answer

answered Mar 23 at 9:22

TiggerTigger

6,61432835

add a comment | 

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0

The syntax is the same, the implementation in your sample is different. For example, the following Javascript:

var x = (1^2)+(3^4)+(5^6)+(4^8);
console.log(x);

Will output x as 25;

The PHP version is the same. For example:

<?=(1^2)+(3^4)+(5^6)+(4^8)?>

Will also render 25.

My guess is you do not have the same values or your syntax is the different in the different languages.

share|improve this answer

answered Mar 23 at 9:22

TiggerTigger

6,61432835

add a comment | 

0

The syntax is the same, the implementation in your sample is different. For example, the following Javascript:

var x = (1^2)+(3^4)+(5^6)+(4^8);
console.log(x);

Will output x as 25;

The PHP version is the same. For example:

<?=(1^2)+(3^4)+(5^6)+(4^8)?>

Will also render 25.

My guess is you do not have the same values or your syntax is the different in the different languages.

share|improve this answer

answered Mar 23 at 9:22

TiggerTigger

6,61432835

add a comment | 

The syntax is the same, the implementation in your sample is different. For example, the following Javascript:

var x = (1^2)+(3^4)+(5^6)+(4^8);
console.log(x);

Will output x as 25;

The PHP version is the same. For example:

<?=(1^2)+(3^4)+(5^6)+(4^8)?>

Will also render 25.

My guess is you do not have the same values or your syntax is the different in the different languages.

share|improve this answer

answered Mar 23 at 9:22

TiggerTigger

6,61432835

var x = (1^2)+(3^4)+(5^6)+(4^8);
console.log(x);

<?=(1^2)+(3^4)+(5^6)+(4^8)?>

share|improve this answer

answered Mar 23 at 9:22

TiggerTigger

6,61432835

answered Mar 23 at 9:22

TiggerTigger

6,61432835

answered Mar 23 at 9:22

TiggerTigger

6,61432835