I have these possible bit flags.
1, 2, 4, 8, 16, 64, 128, 256, 512, 2048, 4096, 16384, 32768, 65536
So each number is like a true/false statement on the server side. So if the first 3 items, and only the first 3 items are marked "true" on the server side, the web service will return a 7. Or if all 14 items above are true, I would still get a single number back from the web service which is is the sum of all those numbers.
What is the best way to handle the number I get back to find out which items are marked as "true"?
Use a bit masking operator. In the C language:
X & 8
is true, if the "8"s bit is set.
You can enumerate the bit masks, and count how many are set.
If it really is the case that the entire word contains bits, and you want to simply
compute how many bits are set, you want in essence a "population count". The absolute
fastest way to get a population count is to execute a native "popcnt" usually
available in your machine's instruction set.
If you don't care about space, you can set up a array countedbits[...] indexed by your value with precomputed bit counts. Then a single memory access computes your bit count.
Often used is just plain "bit twiddling code" that computes bit counts:
(Kernigan's method):
unsigned int v; // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
for (c = 0; v; c++)
{
v &= v - 1; // clear the least significant bit set
}
(parallel bit summming, 32 bits)
v = v - ((v >> 1) & 0x55555555); // reuse input as temporary
v = (v & 0x33333333) + ((v >> 2) & 0x33333333); // temp
c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24; // count
If you haven't seen the bit twiddling hacks before, you're in for a treat.
PHP, being funny, may do funny things with some of this arithmetic.
if (7 & 1) { // if bit 1 is set in returned number (7)
}
Thought the question is old might help someone else. I am putting the numbers in binary as its clearer to understand. The code had not been tested but hope the logic is clear. The code is PHP specific.
define('FLAG_A', 0b10000000000000);
define('FLAG_B', 0b01000000000000);
define('FLAG_C', 0b00100000000000);
define('FLAG_D', 0b00010000000000);
define('FLAG_E', 0b00001000000000);
define('FLAG_F', 0b00000100000000);
define('FLAG_G', 0b00000010000000);
define('FLAG_H', 0b00000001000000);
define('FLAG_I', 0b00000000100000);
define('FLAG_J', 0b00000000010000);
define('FLAG_K', 0b00000000001000);
define('FLAG_L', 0b00000000000100);
define('FLAG_M', 0b00000000000010);
define('FLAG_N', 0b00000000000001);
function isFlagSet($Flag,$Setting,$All=false){
$setFlags = $Flag & $Setting;
if($setFlags and !$All) // at least one of the flags passed is set
return true;
else if($All and ($setFlags == $Flag)) // to check that all flags are set
return true;
else
return false;
}
Usage:
if(isFlagSet(FLAG_A,someSettingsVariable)) // eg: someSettingsVariable = 0b01100000000010
if(isFlagSet(FLAG_A | FLAG_F | FLAG_L,someSettingsVariable)) // to check if atleast one flag is set
if(isFlagSet(FLAG_A | FLAG_J | FLAG_M | FLAG_D,someSettingsVariable, TRUE)) // to check if all flags are set
One way would be to loop through your number, left-shifting it (ie divide by 2) and compare the first bit with 1 using the & operand.
As there is no definite answer with php code, I add this working example:
// returns array of numbers, so for 7 returns array(1,2,4), etc..
function get_bits($decimal) {
$scan = 1;
$result = array();
while ($decimal >= $scan){
if ($decimal & $scan) $result[] = $scan;
$scan<<=1;
}
return $result;
}
Related
I have this code in Qt c++
const unsigned char *packed = reinterpret_cast<const unsigned char*>(data.constData());
res.type = static_cast<int>(packed[0]);
res.period = static_cast<int>(packed[1]);
res.rate = static_cast<qint16>(packed[2] | (packed[3] << 8)) / 100.;
res.edge = static_cast<qint16>(packed[4] | (packed[5] << 8)) / 100.;
return res;
How to convert it from c++ to php
I try this:
$a = unpack ("C*", $data);
$eventList = [];
for ($i=0; $i < $a[1]; $i++)
{
$event = array ();
$index = $i * 6 + 2;
$event["type"] = $a[$index];
$event["period"] = $a[$index+1];
$event["rate"] = ($a[$index+2] | ($a[$index+3] << 8)) / 100;
$event["edge"] = ($a[$index+4] | ($a[$index+5] << 8)) / 100;
}
Edge conver wrong
Very big value.
[edge] => 650.86
must be -4.5
Type, period and rate is good;
Help me please
Don't know the exact answer but some of possible ways to solve the problem:
Check $a[$index+4] and $a[$index+5] value by using var_dump to get its value and type:
var_dump($a[$index+4]);
var_dump($a[$index+5]);
is the data type and its value as expected? Probably good idea is to check as above all data before/after calculation to exactly know what you are dealing with.
Double check your conversion type, perhaps you should't use C* but other, perhaps S or s?
conversion types
If you need type conversion in php you can check how it is done here: Type Juggling and Casting
Note that in PHP you can use a string with ASCII digit that can be treated as digit for calculations:
$foo = 5 * "10 Little Piggies"; // $foo is integer (50)
Which is something you probably don't want.
If you expect negative value but you get positive you have problem because your'e not setting MSB by shifting bits:
The MSB can also correspond to the sign bit of a signed binary number
read-wiki
in case packed[5] should be negative but it isn't
If this not helps then provide data sample and expected values for Edge, Type, period and rate.
I am working in php and I am trying to create 1000 tickets in a database. Each ticket needs it's own unique code that consists of letters and numbers about 6 characters long.
EXP.
Tbl_Tickets
ID code
1 3F2jk7
2 2HGUF1
3 9FJDNJ
4 MFJEY9
5 23988D
I was wondering is there a simple way of doing this with php, or excel, or any other way for that matter. I know that i can use a random number generator, but the check for the Unique would have a large BigO notation and the check would get messy.
Unique is not compatible with random, but the following might suit:
=CHOOSE(RANDBETWEEN(1,2),RANDBETWEEN(0,9),CHAR(RANDBETWEEN(65,90)))
copied across to populate six columns (say A to F) with, in G:
=A1&B1&C1&D1&E1&F1
and both copied down to say row 1100. Then select G, copy Paste Special Values, and Remove Duplicates on ColumnG and select first 1000 entries.
You could easily create an array of strings in php and write it to a database:
function generateRandomString($length = 6, $letters = '1234567890QWERTYUIOPASDFGHJKLZXCVBNM'){
$s = '';
$lettersLength = strlen($letters)-1;
for($i = 0 ; $i < $length ; $i++){
$s .= $letters[rand(0,$lettersLength)];
}
return $s;
}
// Create an array with random strings
for ($i=0; $i<1000; $i++){
$ticket_numbers = array();
$ticket_number = generateRandomString();
while (in_array($ticket_number,$ticket_numbers))
$ticket_number = generateRandomString();
$ticket_numbers[] = $ticket_number;
}
// Write the array to a database
$con = mysqli_connect("myhost","myuser","mypassw","mybd") or die("Error");
foreach ($ticket_numbers as $number){
mysqli_query($con,"Your insert query using the value $number");
}
mysqli_close($con);
This should help you in the right direction though there are probably better ways to do this.
The function generateRandomString() was taken from How to generate random numbers/letters with PHP/Javascript
And another option. Encryption is guaranteed to be unique, so encrypting the numbers 0, 1, 2, ... will give you guaranteed unique random-seeming output. Six characters is 30 bits using Base32, or 36 bits using Base64. You will need a 30 (or 36 bit) cypher. Unless you have a library that includes Hasty Pudding cypher (unlikely) then just implement a simple four round Feistel cypher with the appropriate block size. It will not be completely secure, but it will be enough to defeat casual attacks.
This will produce random strings in column B with no repeats from B1 thru B1001
Sub Lottery()
Dim i As Long, j As Long, c As Collection
Set c = New Collection
v = Split("0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z", ",")
For i = 1 To 5000
can = ""
For j = 1 To 6
can = can & v(Application.RandBetween(0, 35))
Next j
On Error Resume Next
c.Add can, CStr(can)
On Error GoTo 0
If c.Count = 1000 Then Exit For
Next i
For i = 1 To 1000
Cells(i + 1, 2).Value = c(i)
Next i
End Sub
I've got this spot of code that seems it could be done cleaner with pure math (perhaps a logarigthms?). Can you help me out?
The code finds the first power of 2 greater than a given input. For example, if you give it 500, it returns 9, because 2^9 = 512 > 500. 2^8 = 256, would be too small because it's less than 500.
function getFactor($iMaxElementsPerDir)
{
$aFactors = range(128, 1);
foreach($aFactors as $i => $iFactor)
if($iMaxElementsPerDir > pow(2, $iFactor) - 1)
break;
if($i == 0)
return false;
return $aFactors[$i - 1];
}
The following holds true
getFactor(500) = 9
getFactor(1000) = 10
getFactor(2500) = 12
getFactor(5000) = 13
You can get the same effect by shifting the bits in the input to the right and checking against 0. Something like this.
i = 1
while((input >> i) != 0)
i++
return i
The same as jack but shorter. Log with base 2 is the reverse function of 2^x.
echo ceil(log(500, 2));
If you're looking for a "math only" solution (that is a single expression or formula), you can use log() and then take the ceiling value of its result:
$factors = ceil(log(500) / log(2)); // 9
$factors = ceil(log(5000) / log(2)); // 13
I seem to have not noticed that this function accepts a second argument (since PHP 4.3) with which you can specify the base; though internally the same operation is performed, it does indeed make the code shorter:
$factors = ceil(log(500, 2)); // 9
To factor in some inaccuracies, you may need some tweaking:
$factors = floor(log($nr - 1, 2)) + 1;
There are a few ways to do this.
Zero all but the most significant bit of the number, maybe like this:
while (x & x-1) x &= x-1;
and look the answer up in a table. Use a table of length 67 and mod your power of two by 67.
Binary search for the high bit.
If you're working with a floating-point number, inspect the exponent field. This field contains 1023 plus your answer, except in the case where the number is a perfect power of two. You can detect the perfect power case by checking whether the significand field is exactly zero.
If you aren't working with a floating-point number, convert it to floating-point and look at the exponent like in 3. Check for a power of two by testing (x & x-1) == 0 instead of looking at the significand; this is true exactly when x is a power of two.
Note that log(2^100) is the same double as log(nextafter(2^100, 1.0/0.0)), so any solution based on floating-point natural logarithms will fail.
Here's (nonconformant C++, not PHP) code for 4:
int ceillog2(unsigned long long x) {
if (x < 2) return x-1;
double d = x-1;
int ans = (long long &)d >> 52;
return ans - 1022;
}
I wonder if is there a good way to get the number of digits in right/left side of a decimal number PHP. For example:
12345.789 -> RIGHT SIDE LENGTH IS 3 / LEFT SIDE LENGTH IS 5
I know it is readily attainable by helping string functions and exploding the number. I mean is there a mathematically or programmatically way to perform it better than string manipulations.
Your answers would be greatly appreciated.
Update
The best solution for left side till now was:
$left = floor(log10($x))+1;
but still no sufficient for right side.
Still waiting ...
To get the digits on the left side you can do this:
$left = floor(log10($x))+1;
This uses the base 10 logarithm to get the number of digits.
The right side is harder. A simple approach would look like this, but due to floating point numbers, it would often fail:
$decimal = $x - floor($x);
$right = 0;
while (floor($decimal) != $decimal) {
$right++;
$decimal *= 10; //will bring in floating point 'noise' over time
}
This will loop through multiplying by 10 until there are no digits past the decimal. That is tested with floor($decimal) != $decimal.
However, as Ali points out, giving it the number 155.11 (a hard to represent digit in binary) results in a answer of 14. This is because as the number is stored as something like 155.11000000000001 with the 32 bits of floating precision we have.
So instead, a more robust solution is needed. (PoPoFibo's solutions above is particularly elegant, and uses PHPs inherit float comparison functions well).
The fact is, we can never distinguish between input of 155.11 and 155.11000000000001. We will never know which number was originally given. They will both be represented the same. However, if we define the number of zeroes that we can see in a row before we just decide the decimal is 'done' than we can come up with a solution:
$x = 155.11; //the number we are testing
$LIMIT = 10; //number of zeroes in a row until we say 'enough'
$right = 0; //number of digits we've checked
$empty = 0; //number of zeroes we've seen in a row
while (floor($x) != $x) {
$right++;
$base = floor($x); //so we can see what the next digit is;
$x *= 10;
$base *= 10;
$digit = floor($x) - $base; //the digit we are dealing with
if ($digit == 0) {
$empty += 1;
if ($empty == $LIMIT) {
$right -= $empty; //don't count all those zeroes
break; // exit the loop, we're done
}
} else {
$zeros = 0;
}
}
This should find the solution given the reasonable assumption that 10 zeroes in a row means any other digits just don't matter.
However, I still like PopoFibo's solution better, as without any multiplication, PHPs default comparison functions effectively do the same thing, without the messiness.
I am lost on PHP semantics big time but I guess the following would serve your purpose without the String usage (that is at least how I would do in Java but hopefully cleaner):
Working code here: http://ideone.com/7BnsR3
Non-string solution (only Math)
Left side is resolved hence taking the cue from your question update:
$value = 12343525.34541;
$left = floor(log10($value))+1;
echo($left);
$num = floatval($value);
$right = 0;
while($num != round($num, $right)) {
$right++;
}
echo($right);
Prints
85
8 for the LHS and 5 for the RHS.
Since I'm taking a floatval that would make 155.0 as 0 RHS which I think is valid and can be resolved by String functions.
php > $num = 12345.789;
php > $left = strlen(floor($num));
php > $right = strlen($num - floor($num));
php > echo "$left / $right\n";
5 / 16 <--- 16 digits, huh?
php > $parts = explode('.', $num);
php > var_dump($parts);
array(2) {
[0]=>
string(5) "12345"
[1]=>
string(3) "789"
As you can see, floats aren't the easiest to deal with... Doing it "mathematically" leads to bad results. Doing it by strings works, but makes you feel dirty.
$number = 12345.789;
list($whole, $fraction) = sscanf($number, "%d.%d");
This will always work, even if $number is an integer and you’ll get two real integers returned. Length is best done with strlen() even for integer values. The proposed log10() approach won't work for 10, 100, 1000, … as you might expect.
// 5 - 3
echo strlen($whole) , " - " , strlen($fraction);
If you really, really want to get the length without calling any string function here you go. But it's totally not efficient at all compared to strlen().
/**
* Get integer length.
*
* #param integer $integer
* The integer to count.
* #param boolean $count_zero [optional]
* Whether 0 is to be counted or not, defaults to FALSE.
* #return integer
* The integer's length.
*/
function get_int_length($integer, $count_zero = false) {
// 0 would be 1 in string mode! Highly depends on use case.
if ($count_zero === false && $integer === 0) {
return 0;
}
return floor(log10(abs($integer))) + 1;
}
// 5 - 3
echo get_int_length($whole) , " - " , get_int_length($fraction);
The above will correctly count the result of 1 / 3, but be aware that the precision is important.
$number = 1 / 3;
// Above code outputs
// string : 1 - 10
// math : 0 - 10
$number = bcdiv(1, 3);
// Above code outputs
// string : 1 - 0 <-- oops
// math : 0 - INF <-- 8-)
No problem there.
I would like to apply a simple logic.
<?php
$num=12345.789;
$num_str="".$num; // Converting number to string
$array=explode('.',$num_str); //Explode number (String) with .
echo "Left side length : ".intval(strlen($array[0])); // $array[0] contains left hand side then check the string length
echo "<br>";
if(sizeof($array)>1)
{
echo "Left side length : ".intval(strlen($array[1]));// $array[1] contains left hand check the string length side
}
?>
For example, if the pattern is as follows:
bit [10010][1011][1000]
position 54321 4321 4321
result 2 1 4
I want to get the result from right to left position as [2] [1] [4]
If I understand your question correctly, you are looking for a function that returns the index of the least significant 1-bit in an integer. If so, check whether your platform implements the function ffs() ("find first set"). On Linux, you can do man ffs to get the full documentation. On other programming platforms the function may be named differently, e.g. in NVIDIA's CUDA, it exists as a device function __ffs().
Assuming the bit pattern is represented by an int you could do something like
if(bitPattern == 0) {
return 0;
}
int count = 1;
while(bitPattern % 2 == 0) {
bitPattern >>= 1;
count++;
}
return count;
$n = log($x & (~$x+1))/log(2)
~x + 1 is exact the same as -x, as the result of the 2's complement. So why would you use the more complex and slower?
And there are many bithacks to quickly find integer log2x instead of using the much more slower floating point log as above. No slowly divide is needed too. Since x & -x yields only the last bit which is a power of 2, you can use the following function to get log2
unsigned int log2p2(unsigned int v) // 32-bit value to find the log2 of v
{
static const unsigned int b[] = {0xAAAAAAAA, 0xCCCCCCCC, 0xF0F0F0F0,
0xFF00FF00, 0xFFFF0000};
register unsigned int r = (v & b[0]) != 0;
for (i = 4; i > 0; i--) // unroll for speed...
{
r |= ((v & b[i]) != 0) << i;
}
}
There are many other ways to calculate log2x which you can find here
So your code is now simply
log2p2(x & -x);