This question already has answers here:
Is there an easy way to turn an int into an array of ints of each digit?
(11 answers)
Closed 5 years ago.
To sort through 16 million combinations in the 8 Queens problem, I use an 8-digit number in which each digit represents a column, and the value of the digit is the row. No digit may repeat. My function accepts a number, converts it to a string, then to a char array, and then to an int array. Is there a simpler way to arrive at an int array?
Also, are any optimizations available in this code:
private bool IsValidSolution(int MyInt)
{
string sNum = MyInt.ToString();
int iVal = 0;
int i, k, j;
// Eliminate row and column conflicts.
char[] aChars = sNum.ToCharArray();
int[] aInt = Array.ConvertAll(aChars, c => (int)Char.GetNumericValue(c));
// The first false in the array is just a placeholder. Of interest are positions 1 - 8).
bool[] aCheck = new bool[] { false, false, false, false, false, false, false, false, false, false };
for (i = 0; i < 8; i++)
{
iVal = aInt[i];
if (iVal == 0 || iVal == 9) return false;
if (aCheck[iVal] == false)
{
aCheck[iVal] = true;
} else {
return false;
}
}
// Eliminate diagonal conflicts.
for (k = 0; k < 7; k++)
{
j = k + 1;
for (i = j; i < 8; i++)
{
if (Math.Abs(aInt[k] - aInt[i]) == Math.Abs(k - i)) return false;
}
}
return true;
}
Yes you can, using some arithmetic on the number let's say n = 4,732
To get the thousands digit you can do (int) n/1000.
To get the hundreds digit you do: ((int) n/100) % 10
The tens digit (3) you do: ((int) n/10) %10
and so on... and place each value in a block of the array of ints
Can be done in a loop
Related
I need to validate ICCID, I found only one algo:
int numberStringLength = 18;
int cs = 0;
int dodd;
for (int i = 0; i < numberStringLength; i += 2)
{
dodd = Convert.ToInt32(iccid.Substring(i + 1, 1)) << 1;
cs += Convert.ToInt32(iccid.Substring(i, 1)) + (int)(dodd / 10) + (dodd % 10);
}
cs = (10-(cs % 10)) % 10;
if (cs == Convert.ToInt32(iccid.Substring(numberStringLength, 1)))
{
return true;
}
else
{
return false;
}
but it returns false for 100% right ICCID (89148000005339755555). Where can I get real ICCID algo?
Thanks
According to Wikipedia, ICCIDs use the Luhn algorithm.
Your code that you found is a bit broken, as it assumes that the value has an odd number of digits (an even number of normal digits, plus 1 check digit). It starts parsing the value from the left-most digit, and assumes that this left-most digit ("8" in your example) is not doubled and the next one ("9") is doubled. But this is not correct if the value has an even number of digits. The "8" should be the one that's doubled in your case.
Thankfully, it's very easy to implement the Luhn algorithm ourselves, properly, using that Wikipedia page as reference:
string input = "89148000005339755555";
int sum = 0;
// We'll use index i = 0 means the right-most digit, i = 1 is second-right, etc
for (int i = 0; i < input.Length; i++)
{
// Get the digit at the i'th position from the right
int digit = int.Parse(input[input.Length - i - 1].ToString());
// If it's in an odd position (starting from the right), then double it.
if (i % 2 == 1)
{
digit *= 2;
// If it's now >= 10, subtract 9
if (digit >= 10)
{
digit -= 9;
}
}
sum += digit;
}
// It's a pass if the result is a multiple of 10
bool pass = sum % 10 == 0;
Console.WriteLine(pass ? "Pass" : "Fail");
See it on dotnetfiddle.net.
This question already has answers here:
how can read values and put them in Array C#
(3 answers)
Closed 3 years ago.
I have been messing around with Lambda expressions in C#, teaching myself, and trying to test myself here. The problem with my code is not in evaluating the array for these set conditions, but rather I am having a difficult time capturing user input to build out the array. Have tried a few different methods, some mentioned here, but seem to have trouble adapting them to my code. Also any input on simplifying my expressions for checking if it has odd only or even only would be appreciated! I feel they are slightly bloated. Thanks!
using System;
namespace BuildingArrays
{
class Program
{
static void Main(string[] args)
{
int[] numbers = {Int32.TryParse(string, Console.ReadLine())};
bool hasOddOnly = Array.Exists(numbers, (int num) =>
{
bool hasEven = num % 2 == 0;
return hasEven;
});
hasOddOnly = !hasOddOnly;
bool hasEvenOnly = Array.Exists(numbers, (int num) =>
{
bool hasOdd = num % 2 != 0;
return hasOdd;
});
hasEvenOnly = !hasEvenOnly;
bool hasOddAnd4 = Array.Exists(numbers, (int num) =>
{
bool hasOdd = num % 2 != 0;
bool is4 = num == 4;
return is4 && hasOdd;
});
bool multipleOf4 = Array.Exists(numbers, (int num)=>
{
bool multiple = num % 4 == 0;
bool multiple2 = num % 3 == 0;
return multiple || multiple2;
});
bool multipleOf4and3 = Array.Exists(numbers, (int num)=>
{
bool multipleBoth = num % 4 == 0 && num % 3 == 0;
return multipleBoth;
});
Console.WriteLine($"This array contains odd numbers and 4 is {hasOddAnd4}");
Console.WriteLine($"This array contains ONLY odd numbers is {hasOddOnly}.");
Console.WriteLine($"This array contains ONLY even numbers is {hasEvenOnly}.");
Console.WriteLine($"This array contains either a multiple of 4, or a multiple of 3 is {multipleOf4}.");
Console.WriteLine($"This array contains a number which is both a multiple of 4 and 3 is {multipleOf4and3}");
}
}
}
Array.Exists takes a Predicate<T> as the second parameter. That should look like this:
bool hasEven = Array.Exists(numbers, x => x % 2 == 0);
To fill your array, you're going to need something like this:
for (int i=0; i< numbers.Length; i++)
numbers[i] = Int32.Parse(Console.ReadLine());
Further Reading
Array.Exists on MSDN
How to fill an array from user input in C#
I try to write program that check the ratio between odd and even
digits in a given number. I've had some problems with this code:
static void Main(string[] args)
{
int countEven = 0 ;
int countOdd = 0 ;
Console.WriteLine("insert a number");
int num = int.Parse(Console.ReadLine());
int length = num.GetLength;
for (int i = 0;i<length ; i++)
{
if((num/10)%2) == 0)
int countEven++;
}
}
any ideas?
The problem is that int does not have a length, only the string representation of it has one.As an alternative to m.rogalski answer, you can treat the input as a string to get all the digits one by one. Once you have a digit, then parsing it to int and checking if it is even or odd is trivial.Would be something like this:
int countEven = 0;
int countOdd = 0;
Console.WriteLine("insert a number");
string inputString = Console.ReadLine();
for (int i = 0; i < inputString.Length; i++)
{
if ((int.Parse(inputString[i].ToString()) % 2) == 0)
countEven++;
else
countOdd++;
}
Linq approach
Console.WriteLine("insert a number");
string num = Console.ReadLine(); // check for valid number here?
int countEven = num.Select(x => x - '0').Count(x => x % 2 == 0);
int countOdd = num.Select(x => x - '0').Count(x => x % 2 != 0);
Let's assume your input is : 123456
Now all you have to do is to get the modulo from the division by ten : int m = num % 10;
After that just check if bool isEven = m % 2 == 0;
On the end you have to just divide your input number by 10 and repeat the whole process till the end of numbers.
int a = 123456, oddCounter = 0, evenCounter = 0;
do
{
int m = a % 10;
switch(m % 2)
{
case 0:
evenCounter++;
break;
default: // case 1:
oddCounter++;
break;
}
//bool isEven = m % 2 == 0;
}while( ( a /= 10 ) != 0 );
Online example
Made a small change to your code and it works perfectly
int countEven = 0;
int countOdd = 0;
Console.WriteLine( "insert a number" );
char[] nums = Console.ReadLine().ToCharArray();
for ( int i = 0; i < nums.Length; i++ )
{
if ( int.Parse( nums[i].ToString() ) % 2 == 0 )
{
countEven++;
}
else
{
countOdd++;
}
}
Console.WriteLine($"{countEven} even numbers \n{countOdd} odd numbers");
Console.ReadKey();
What I do is get each number as a a character in an array char[] and I loop through this array and check if its even or not.
If the Input number is a 32-bit integer (user pick the length of the number)
if asked:
The number of even digits in the input number
Product of odd digits in the input number
The sum of all digits of the input number
private void button1_Click(object sender, EventArgs e) {
int num = ConvertToInt32(textBox1.Text);
int len_num = textBox1.Text.Length;
int[] arn = new int[len_num];
int cEv = 0; pOd = 0; s = 0;
for (int i = len_num-1; i >= 0; i--) { // loop until integer length is got down to 1
arn[i] = broj % 10; //using the mod we put the last digit into a declared array
if (arn[i] % 2 == 0) { // then check, is current digit even or odd
cEv++; // count even digits
} else { // or odd
if (pOd == 0) pOd++; // avoid product with zero
pOd *= arn [i]; // and multiply odd digits
}
num /= 10; // we divide by 10 until it's length is get to 1(len_num-1)
s += arn [i]; // sum of all digits
}
// and at last showing it in labels...
label2.Text = "a) The even digits count is: " + Convert.ToString(cEv);
label3.Text = "b) The product of odd digits is: " + Convert.ToString(pOd);
label4.Text = "c) The sum of all digits in this number is: " + Convert.ToString(s);
}
All we need in the interface is the textbox for entering the number, the button for the tasks, and labels to show obtained results. Of course, we have the same result if we use a classic form for the for loop like for (int i = 0; and <= len_num-1; i++) - because the essence is to count the even or odd digits rather than the sequence of the digits entry into the array
static void Main(string args[]) {
WriteLine("Please enter a number...");
var num = ReadLine();
// Check if input is a number
if (!long.TryParse(num, out _)) {
WriteLine("NaN!");
return;
}
var evenChars = 0;
var oddChars = 0;
// Convert string to char array, rid of any non-numeric characters (e.g.: -)
num.ToCharArray().Where(c => char.IsDigit(c)).ToList().ForEach(c => {
byte.TryParse(c.ToString(), out var b);
if (b % 2 == 0)
evenChars++;
else
oddChars++;
});
// Continue with code
}
EDIT:
You could also do this with a helper (local) function within the method body:
static void Main(string args[]) {
WriteLine("Please enter a number...");
var num = ReadLine();
// Check if input is a number
if (!long.TryParse(num, out _)) {
WriteLine("NaN!");
return;
}
var evenChars = 0;
var oddChars = 0;
// Convert string to char array, rid of any non-numeric characters (e.g.: -)
num.ToCharArray().Where(c => char.IsDigit(c)).ToList().ForEach(c => {
byte.TryParse(c.ToString(), out var b);
if (b % 2 == 0)
evenChars++;
else
oddChars++;
// Alternative method:
IsEven(b) ? evenChars++ : oddChars++;
});
// Continue with code
bool IsEven(byte b) => b % 2 == 0;
}
Why am I using a byte?
Dealing with numbers, it is ideal to use datatypes that don't take up as much RAM.
Granted, not as much an issue nowadays with multiple 100s of gigabytes possible, however, it is something not to be neglected.
An integer takes up 32 bits (4 bytes) of RAM, whereas a byte takes up a single byte (8 bits).
Imagine you're processing 1 mio. single-digit numbers, and assigning them each to integers. You're using 4 MiB of RAM, whereas the byte would only use up 1 MiB for 1 mio. numbers.
And seeming as a single-digit number (as is used in this case) can only go up to 9 (0-9), you're wasting a potential of 28 bits of memory (2^28) - whereas a byte can only go up to 255 (0-255), you're only wasting a measly four bits (2^4) of memory.
At first user gives a number (n) to program, for example 5.
the program must find the smallest number that can be divided to n (5).
and this number can only consist of digits 0 and 9 not any other digits.
for example if user gives 5 to program.
numbers that can be divided to 5 are:
5, 10, 15, 20, 25, 30, ..., 85, 90, 95, ...
but 90 here is the smallest number that can be divided to 5 and also consist of digits (0 , 9). so answer for 5 must be 90.
and answer for 9 is 9, because it can be divided to 9 and consist of digit (9).
my code
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
Int64 i ,j=1,y=x;
bool t = false;
for (i = x + 1; t == false; i++)
{
if (i % 9 == 0 && i % 10 == 0 && i % x == 0)
{
j = i;
for (; (i /= 10) != 0; )
{
i /= 10;
if (i == 0)
t = true;
continue;
}
}
}
lblAnswer.Text = Convert.ToString(j);
If you're happy to go purely functional then this works:
Func<IEnumerable<long>> generate = () =>
{
Func<long, IEnumerable<long>> extend =
x => new [] { x * 10, x * 10 + 9 };
Func<IEnumerable<long>, IEnumerable<long>> generate2 = null;
generate2 = ns =>
{
var clean = ns.Where(n => n > 0).ToArray();
return clean.Any()
? clean.Concat(generate2(clean.SelectMany(extend)))
: Enumerable.Empty<long>();
};
return generate2(new[] { 9L, });
};
Func<long, long?> f = n =>
generate()
.Where(x => x % n == 0L)
.Cast<long?>()
.FirstOrDefault();
So rather than iterate through all possible values and test for 0 & 9 and divisibility, this just generates only numbers with 0 & 9 and then only tests for visibility. It is much faster this way.
I can call it like this:
var result = f(5L); // 90L
result = f(23L); //990909L
result = f(123L); //99999L
result = f(12321L); //90900999009L
result = f(123212L); //99909990090000900L
result = f(117238L); //990990990099990990L
result = f(1172438L); //null == No answer
These results are super fast. f(117238L) returns a result on my computer in 138ms.
You can try this way :
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
int counter;
for (counter = 1; !isValid(x * counter); counter++)
{
}
lblAnswer.Text = Convert.ToString(counter*x);
code above works by searching multiple of x incrementally until result that satisfy criteria : "consist of only 0 and or 9 digits" found. By searching only multiple of x, it is guaranteed to be divisible by x. So the rest is checking validity of result candidate, in this case using following isValid() function :
private static bool isValid(int number)
{
var lastDigit = number%10;
//last digit is invalid, return false
if (lastDigit != 0 & lastDigit != 9) return false;
//last digit is valid, but there is other digit(s)
if(number/10 >= 1)
{
//check validity of digit(s) before the last
return isValid(number/10);
}
//last digit is valid, and there is no other digit. return true
return true;
}
About strange empty for loop in snippet above, it is just syntactic sugar, to make the code a bit shorter. It is equal to following while loop :
counter = 1;
while(!isValid(input*counter))
{
counter++;
}
Use this simple code
int inputNumber = 5/*Or every other number, you can get this number from input.*/;
int result=1;
for (int i = 1; !IsOk(result,inputNumber); i++)
{
result = i*inputNumber;
}
Print(result);
IsOk method is here:
bool IsOk(int result, int inputNumber)
{
if(result%inputNumber!=0)
return false;
if(result.ToString().Replace("9",string.Empty).Replace("0",string.Empty).Length!=0)
return false;
return true;
}
My first solution has very bad performance, because of converting a number to string and looking for characters '9' and '0'.
New solution:
My new solution has very good performance and is a technical approach since of using Breadth-first search(BFS).
Algorithm of this solution:
For every input number, test 9, if it is answer print it, else add 2 child numbers (90 & 99) to queue, and continue till finding answer.
int inputNumber = 5;/*Or every other number, you can get this number from input.*/
long result;
var q = new Queue<long>();
q.Enqueue(9);
while (true)
{
result = q.Dequeue();
if (result%inputNumber == 0)
{
Print(result);
break;
}
q.Enqueue(result*10);
q.Enqueue(result*10 + 9);
}
Trace of number creation:
9
90,99
900,909,990,999
9000,9009,9090,9099,9900,9909,9990,9999
.
.
.
I wrote this code for console, and i used goto command however it is not prefered but i could not write it with only for.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace main
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Enter your number");
Int64 x = Convert.ToInt64(Console.ReadLine());
Int64 y, j, i, k, z = x;
x = x + 5;
loop:
x++;
for (i = 0, y = x; y != 0; i++)
y /= 10;
for (j = x, k = i; k != 0; j /= 10, k--)
{
if (j % 10 != 9)
if (j % 10 != 0)
goto loop;
}
if (x % z != 0)
goto loop;
Console.WriteLine("answer:{0}",x);
Console.ReadKey();
}
}
}
I have a datastructure with a field of the float-type. A collection of these structures needs to be sorted by the value of the float. Is there a radix-sort implementation for this.
If there isn't, is there a fast way to access the exponent, the sign and the mantissa.
Because if you sort the floats first on mantissa, exponent, and on exponent the last time. You sort floats in O(n).
Update:
I was quite interested in this topic, so I sat down and implemented it (using this very fast and memory conservative implementation). I also read this one (thanks celion) and found out that you even dont have to split the floats into mantissa and exponent to sort it. You just have to take the bits one-to-one and perform an int sort. You just have to care about the negative values, that have to be inversely put in front of the positive ones at the end of the algorithm (I made that in one step with the last iteration of the algorithm to save some cpu time).
So heres my float radixsort:
public static float[] RadixSort(this float[] array)
{
// temporary array and the array of converted floats to ints
int[] t = new int[array.Length];
int[] a = new int[array.Length];
for (int i = 0; i < array.Length; i++)
a[i] = BitConverter.ToInt32(BitConverter.GetBytes(array[i]), 0);
// set the group length to 1, 2, 4, 8 or 16
// and see which one is quicker
int groupLength = 4;
int bitLength = 32;
// counting and prefix arrays
// (dimension is 2^r, the number of possible values of a r-bit number)
int[] count = new int[1 << groupLength];
int[] pref = new int[1 << groupLength];
int groups = bitLength / groupLength;
int mask = (1 << groupLength) - 1;
int negatives = 0, positives = 0;
for (int c = 0, shift = 0; c < groups; c++, shift += groupLength)
{
// reset count array
for (int j = 0; j < count.Length; j++)
count[j] = 0;
// counting elements of the c-th group
for (int i = 0; i < a.Length; i++)
{
count[(a[i] >> shift) & mask]++;
// additionally count all negative
// values in first round
if (c == 0 && a[i] < 0)
negatives++;
}
if (c == 0) positives = a.Length - negatives;
// calculating prefixes
pref[0] = 0;
for (int i = 1; i < count.Length; i++)
pref[i] = pref[i - 1] + count[i - 1];
// from a[] to t[] elements ordered by c-th group
for (int i = 0; i < a.Length; i++){
// Get the right index to sort the number in
int index = pref[(a[i] >> shift) & mask]++;
if (c == groups - 1)
{
// We're in the last (most significant) group, if the
// number is negative, order them inversely in front
// of the array, pushing positive ones back.
if (a[i] < 0)
index = positives - (index - negatives) - 1;
else
index += negatives;
}
t[index] = a[i];
}
// a[]=t[] and start again until the last group
t.CopyTo(a, 0);
}
// Convert back the ints to the float array
float[] ret = new float[a.Length];
for (int i = 0; i < a.Length; i++)
ret[i] = BitConverter.ToSingle(BitConverter.GetBytes(a[i]), 0);
return ret;
}
It is slightly slower than an int radix sort, because of the array copying at the beginning and end of the function, where the floats are bitwise copied to ints and back. The whole function nevertheless is again O(n). In any case much faster than sorting 3 times in a row like you proposed. I dont see much room for optimizations anymore, but if anyone does: feel free to tell me.
To sort descending change this line at the very end:
ret[i] = BitConverter.ToSingle(BitConverter.GetBytes(a[i]), 0);
to this:
ret[a.Length - i - 1] = BitConverter.ToSingle(BitConverter.GetBytes(a[i]), 0);
Measuring:
I set up some short test, containing all special cases of floats (NaN, +/-Inf, Min/Max value, 0) and random numbers. It sorts exactly the same order as Linq or Array.Sort sorts floats:
NaN -> -Inf -> Min -> Negative Nums -> 0 -> Positive Nums -> Max -> +Inf
So i ran a test with a huge array of 10M numbers:
float[] test = new float[10000000];
Random rnd = new Random();
for (int i = 0; i < test.Length; i++)
{
byte[] buffer = new byte[4];
rnd.NextBytes(buffer);
float rndfloat = BitConverter.ToSingle(buffer, 0);
switch(i){
case 0: { test[i] = float.MaxValue; break; }
case 1: { test[i] = float.MinValue; break; }
case 2: { test[i] = float.NaN; break; }
case 3: { test[i] = float.NegativeInfinity; break; }
case 4: { test[i] = float.PositiveInfinity; break; }
case 5: { test[i] = 0f; break; }
default: { test[i] = test[i] = rndfloat; break; }
}
}
And stopped the time of the different sorting algorithms:
Stopwatch sw = new Stopwatch();
sw.Start();
float[] sorted1 = test.RadixSort();
sw.Stop();
Console.WriteLine(string.Format("RadixSort: {0}", sw.Elapsed));
sw.Reset();
sw.Start();
float[] sorted2 = test.OrderBy(x => x).ToArray();
sw.Stop();
Console.WriteLine(string.Format("Linq OrderBy: {0}", sw.Elapsed));
sw.Reset();
sw.Start();
Array.Sort(test);
float[] sorted3 = test;
sw.Stop();
Console.WriteLine(string.Format("Array.Sort: {0}", sw.Elapsed));
And the output was (update: now ran with release build, not debug):
RadixSort: 00:00:03.9902332
Linq OrderBy: 00:00:17.4983272
Array.Sort: 00:00:03.1536785
roughly more than four times as fast as Linq. That is not bad. But still not yet that fast as Array.Sort, but also not that much worse. But i was really surprised by this one: I expected it to be slightly slower than Linq on very small arrays. But then I ran a test with just 20 elements:
RadixSort: 00:00:00.0012944
Linq OrderBy: 00:00:00.0072271
Array.Sort: 00:00:00.0002979
and even this time my Radixsort is quicker than Linq, but way slower than array sort. :)
Update 2:
I made some more measurements and found out some interesting things: longer group length constants mean less iterations and more memory usage. If you use a group length of 16 bits (only 2 iterations), you have a huge memory overhead when sorting small arrays, but you can beat Array.Sort if it comes to arrays larger than about 100k elements, even if not very much. The charts axes are both logarithmized:
(source: daubmeier.de)
There's a nice explanation of how to perform radix sort on floats here:
http://www.codercorner.com/RadixSortRevisited.htm
If all your values are positive, you can get away with using the binary representation; the link explains how to handle negative values.
By doing some fancy casting and swapping arrays instead of copying this version is 2x faster for 10M numbers as Philip Daubmeiers original with grouplength set to 8. It is 3x faster as Array.Sort for that arraysize.
static public void RadixSortFloat(this float[] array, int arrayLen = -1)
{
// Some use cases have an array that is longer as the filled part which we want to sort
if (arrayLen < 0) arrayLen = array.Length;
// Cast our original array as long
Span<float> asFloat = array;
Span<int> a = MemoryMarshal.Cast<float, int>(asFloat);
// Create a temp array
Span<int> t = new Span<int>(new int[arrayLen]);
// set the group length to 1, 2, 4, 8 or 16 and see which one is quicker
int groupLength = 8;
int bitLength = 32;
// counting and prefix arrays
// (dimension is 2^r, the number of possible values of a r-bit number)
var dim = 1 << groupLength;
int groups = bitLength / groupLength;
if (groups % 2 != 0) throw new Exception("groups must be even so data is in original array at end");
var count = new int[dim];
var pref = new int[dim];
int mask = (dim) - 1;
int negatives = 0, positives = 0;
// counting elements of the 1st group incuding negative/positive
for (int i = 0; i < arrayLen; i++)
{
if (a[i] < 0) negatives++;
count[(a[i] >> 0) & mask]++;
}
positives = arrayLen - negatives;
int c;
int shift;
for (c = 0, shift = 0; c < groups - 1; c++, shift += groupLength)
{
CalcPrefixes();
var nextShift = shift + groupLength;
//
for (var i = 0; i < arrayLen; i++)
{
var ai = a[i];
// Get the right index to sort the number in
int index = pref[( ai >> shift) & mask]++;
count[( ai>> nextShift) & mask]++;
t[index] = ai;
}
// swap the arrays and start again until the last group
var temp = a;
a = t;
t = temp;
}
// Last round
CalcPrefixes();
for (var i = 0; i < arrayLen; i++)
{
var ai = a[i];
// Get the right index to sort the number in
int index = pref[( ai >> shift) & mask]++;
// We're in the last (most significant) group, if the
// number is negative, order them inversely in front
// of the array, pushing positive ones back.
if ( ai < 0) index = positives - (index - negatives) - 1; else index += negatives;
//
t[index] = ai;
}
void CalcPrefixes()
{
pref[0] = 0;
for (int i = 1; i < dim; i++)
{
pref[i] = pref[i - 1] + count[i - 1];
count[i - 1] = 0;
}
}
}
You can use an unsafe block to memcpy or alias a float * to a uint * to extract the bits.
I think your best bet if the values aren't too close and there's a reasonable precision requirement, you can just use the actual float digits before and after the decimal point to do the sorting.
For example, you can just use the first 4 decimals (be they 0 or not) to do the sorting.