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• Omniauth + Pow Issue

  - by neon

  I am having a strange issue with Pow and Omniauth. Omniauth (Facebook Login) works fine when using localhost:3000, but when using Pow (appname.dev) things get fishy. Users are taken through the redirect and properly created if they don't exist in the database, as they should be. After this, however, they are redirected to the root_path and not signed in. Their record is saved in the database as expected, but sign in does not occur. Again, this is only happening on Pow (and lvh.me), and not on localhost. Any ideas? I am using the Devise/Omniauth approach for sign-in, and the controller code looks like this: def facebook @user = User.find_for_facebook_oauth(request.env["omniauth.auth"], current_user) if @user.persisted? flash[:notice] = I18n.t "devise.omniauth_callbacks.success", :kind => "Facebook" sign_in_and_redirect @user, :event => :authentication else session["devise.facebook_data"] = request.env["omniauth.auth"] redirect_to new_user_registration_url end end Again, the user is persisted but there is no flash notice or sign_in that occurs when using POW.

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• java.math.BigInteger pow(exponent) question

  - by Jan Kraus

  Hi, I did some tests on pow(exponent) method. Unfortunately, my math skills are not strong enough to handle the following problem. I'm using this code: BigInteger.valueOf(2).pow(var); Results: var | time in ms 2000000 | 11450 2500000 | 12471 3000000 | 22379 3500000 | 32147 4000000 | 46270 4500000 | 31459 5000000 | 49922 See? 2,500,000 exponent is calculated almost as fast as 2,000,000. 4,500,000 is calculated much faster then 4,000,000. Why is that? To give you some help, here's the original implementation of BigInteger.pow(exponent): public BigInteger pow(int exponent) { if (exponent < 0) throw new ArithmeticException("Negative exponent"); if (signum==0) return (exponent==0 ? ONE : this); // Perform exponentiation using repeated squaring trick int newSign = (signum<0 && (exponent&1)==1 ? -1 : 1); int[] baseToPow2 = this.mag; int[] result = {1}; while (exponent != 0) { if ((exponent & 1)==1) { result = multiplyToLen(result, result.length, baseToPow2, baseToPow2.length, null); result = trustedStripLeadingZeroInts(result); } if ((exponent >>>= 1) != 0) { baseToPow2 = squareToLen(baseToPow2, baseToPow2.length, null); baseToPow2 = trustedStripLeadingZeroInts(baseToPow2); } } return new BigInteger(result, newSign); }

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• BigInteger.pow(BigInteger) ?

  - by PeterW

  I'm playing with numbers in Java, and want to see how big a number I can make. It is my understanding that BigInteger can hold a number of infinite size, so long as my computer has enough Memory to hold such a number, correct? My problem is that BigInteger.pow accepts only an int, not another BigInteger, which means I can only use a number up to 2,147,483,647 as the exponent. Is it possible to use the BigInteger class as such? BigInteger.pow(BigInteger) Thanks.

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• ruby-debug with Pow -- breakpoints never hit

  - by 99miles

  I'm trying to use ruby-debug with Pow. Rails 3 app. I have done everything here: https://gist.github.com/1098830 I've restarted the server and machine several times. I can get rdebug to connect: ? rdebug -c Connected. but it never stops at the breakpoints. Any idea what could be going on? I got it to hit a few breakpoints a few hours ago, and not since. controller def index debugger ... end Gemfile gem 'ruby-debug19', :require => 'ruby-debug' development.rb EG::Application.configure do ... require 'ruby-debug' Debugger.start_remote Debugger.settings[:autoeval] = true puts "=> Debugger enabled" end

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• Alternatives to Firefox 3.5/POW on ubuntu natty to use SSJS

  - by Juan Sebastian Totero

  I have to install FFX 3.5 on my 11.04 machine. It's needed because I'm helping a friend of mine in a project involving Server-Side Javascript, and he is using POW webserver, actually avaliable for Linux only as a Firefox AddOn. (I know it's a dumb thing) The addon is compatible only with FFX 3.5 and older, but I cant' find any official package of Firefox 3.5 for linux. So the questions are two: Where can i find a package of Firefox 3.5 for linux? Is there any alternative SSJS webserver out there? IT's main use will be displaying ssjs files in the browser (possibly on-the-fly, that means I have not to create a webserver in SSJS, like in the case of nodejs)

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• C2244 when trying to call the pow function from inline assembly

  - by schrödingers cat

  I would like to call the pow function from inline assembly. The problem is i'm getting error C2244: 'pow' : unable to match function definition to an existing declaration. I'm new to assembly so this may be a trivial question but how do i resolve this? I guess it has something to do with the compiler not beeing able to properly resolve the overload of pow. The following code fragment is causing the error: do_POW: // push first argument to the stack sub esp, size value_type fld qword ptr [ecx] fstp qword ptr [esp] // push second argument to the stack sub esp, size value_type fld qword ptr [ecx - size value_type] fstp qword ptr [esp]and pop fpu stack // call the pow function call pow sub ecx, size value_type fstp qword ptr [ecx] add esp, 2 * size value_type jmp loop_start

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• Polynomial division overloading operator (solved)

  - by Vlad

  Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) const { polinom Result, temp2; polinom temp = *this; Iter i = temp.poly.begin(); constIter j = P.poly.begin(); int resultSize = 0; if (temp.poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); temp = temp - Result * P; } else { Result.insert(0, 0); } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); if (Result.poly.size() < 2) temp2 = Result; else { temp2 = Result; resultSize = Result.poly.size(); for (int k = 1 ; k != resultSize; k++) temp2.poly.pop_front(); } temp = temp - temp2 * P; } else break; } } return Result; } }; The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Final Update After listening to Dave, I finally made it by overloading both / and & to return the quotient and the remainder so thanks a lot everyone for your help and especially you Dave for your great idea! P.S. If anyone wants for me to post these 2 overloaded operator please ask it by commenting on my post (and maybe give a vote up for everyone involved).

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• Polynomial division overloading operator

  - by Vlad

  Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) { polinom Result, temp; Iter i = poly.begin(); constIter j = P.poly.begin(); if (poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); *this = *this - Result; } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); temp = Result * P; *this = *this - temp; } else break; } } return Result; } The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Update Because no one seems to understand how i thought the algorithm, i'll explain: If the dividend contains only one term, we simply insert the quotient in Result, then we multiply it with the divisor ans subtract it from the first polynomial which stores the remainder. If the polynomial we do this until the second polynomial( P in this case) becomes bigger. I think this algorithm is called long division, isn't it? So based on these, can anyone help me with overloading the / operator correctly for my class? Thanks!

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• Performance of std::pow - cache misses???

  - by Eamon Nerbonne

  I've been trying to optimize a numeric program of mine, and have run into something of a mystery. I'm looping over code that performs thousands of floating point operations, and just 1 call to pow nevertheless, that call takes 5% of the time... That's not necessarily a critical issue, but it is odd, so I'd like to understand what's happening. When I profiled for cache misses, VS.NET 2010RC's profiler reports that virtually all cache misses are occurring in std::pow... so... what's up with that? Is there a faster alternative? I tried powf, but that's only slightly faster; it's still responsible for an abnormal number of cache misses. Why would a basic function like pow cause cache-misses?

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• pow doesn't accept the second parameter to be a variable on gcc

  - by Daziplqa

  pow doesn't accept the second parameter to be a variable on gcc The following code works fine on VC++10 // file test.cc #include "stdafx.h" #include <stdio.h> #include <math.h> int main(void) { double x = 10; int y = 20; printf("%f\n", pow(x, y)); return 0; } But the following code doesn't not work on gcc: // test.c #include <stdio.h> #include <math.h> int main(void) { double x = 10; int y = 20; printf("%f\n", pow(x, y)); // error here, says no such function, however when pass the second argument in `pow` for the code runs by gcc, It works fine! return 0; }

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• The most efficient way of implementing pow() function in floating point

  - by user319873

  Hi, I am trying to implement my own version of pow() and sqrt() function as my custom library doesn't have pow()/sqrt() floating point support. Can anyone help?

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• C++ Question on the pow function.

  - by Sagistic

  I'm trying to get this expression to work, I'm pretty sure its not the parenthesis because I counted all of them. Perhaps there something I'm doing wrong involving the parameter pow (x,y). double calculatePeriodicPayment() { periodicPaymentcalc = (loan * ((interestRate / yearlyPayment))) / (1-((pow ((1+(interestRate / yearlyPayment)))),(-(yearlyPayment * numOfYearLoan)))); return periodicPaymentcalc; }

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• Tail-recursive pow() algorithm with memoization?

  - by Dan

  I'm looking for an algorithm to compute pow() that's tail-recursive and uses memoization to speed up repeated calculations. Performance isn't an issue; this is mostly an intellectual exercise - I spent a train ride coming up with all the different pow() implementations I could, but was unable to come up with one that I was happy with that had these two properties. My best shot was the following: def calc_tailrec_mem(base, exp, cache_line={}, acc=1, ctr=0): if exp == 0: return 1 elif exp == 1: return acc * base elif exp in cache_line: val = acc * cache_line[exp] cache_line[exp + ctr] = val return val else: cache_line[ctr] = acc return calc_tailrec_mem(base, exp-1, cache_line, acc * base, ctr + 1) It works, but it doesn't memorize the results of all calculations - only those with exponents 1..exp/2 and exp.

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• Implementing Math.pow() in blackberry

  - by Rahul Varma

  Hi, I cant find any pre defined function in blackberry JDE to implement Math.pow(). Can anyone please suggest how to power the values that are retrieved from the Edit text fields(Like X^Y value)...

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• Using pow() for large number

  - by g4ur4v

  I am trying to solve a problem, a part of which requires me to calculate (2^n)%1000000007 , where n<=10^9. But my following code gives me output "0" even for input like n=99. Is there anyway other than having a loop which multilplies the output by 2 every time and finding the modulo every time (this is not I am looking for as this will be very slow for large numbers). #include<stdio.h> #include<math.h> #include<iostream> using namespace std; int main() { unsigned long long gaps,total; while(1) { cin>>gaps; total=(unsigned long long)powf(2,gaps)%1000000007; cout<<total<<endl; } }

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• Use Math class to calculate

  - by LC

  Write a JAVA program that calculates the surface area of a cylinder, C= 2( p r2) + 2 p r h, where r is the radius and h is the height of the cylinder. Allow the user to enter in a radius and a height. Format the output to three decimal places. Use the constant PI and the method pow() from the Math class. This is what I've done so far but it can not run. Can anyone help me ? import Java.util Scanner; public class Ex5 { public static void main (String[] args) { Scanner input.new Scanner(System.in); double radius,height,area; System.out.print("Please enter the radius of the Cylinder"); radius = input.nextDouble(); System.out.print("Please enter the height of the Cylinder"); height = input.nextDouble(); area = (4*Math.PI*radius)+2*Math.PIMath.POW(radius,2); System.out.println("The surface of the cylinder" + area); } }

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• OpenCL - incremental summation during compute

  - by user1721997

  I'm absolutelly novice in OpenCL programming. For my app. (molecular simulaton) I wrote a kernel for calculate intermolecular potential of lennard-jones liquid. In this kernel I need to compute cumulative value of the potential of all particles with one: __kernel void Molsim(__global const float* inmatrix, __global float* fi, const int c, const float r1, const float r2, const float r3, const float rc, const float epsilon, const float sigma, const float h1, const float h23) { float fi0; float fi1; float d; unsigned int i = get_global_id(0); //number of particles (typically 2000) if(c!=i) { // potential before particle movement d=sqrt(pow((0.5*h1-fabs(0.5*h1-fabs(inmatrix[c*3]-inmatrix[i*3]))),2.0)+pow((0.5*h23-fabs(0.5*h23-fabs(inmatrix[c*3+1]-inmatrix[i*3+1]))),2.0)+pow((0.5*h23-fabs(0.5*h23-fabs(inmatrix[c*3+2]-inmatrix[i*3+2]))),2.0)); if(d<rc) { fi0=4.0*epsilon*(pow(sigma/d,12.0)-pow(sigma/d,6.0)); } else { fi0=0; } // potential after particle movement d=sqrt(pow((0.5*h1-fabs(0.5*h1-fabs(r1-inmatrix[i*3]))),2.0)+pow((0.5*h23-fabs(0.5*h23-fabs(r2-inmatrix[i*3+1]))),2.0)+pow((0.5*h23-fabs(0.5*h23-fabs(r3-inmatrix[i*3+2]))),2.0)); if(d<rc) { fi1=4.0*epsilon*(pow(sigma/d,12.0)-pow(sigma/d,6.0)); } else { fi1=0; } // cumulative difference of potentials fi[0]+=fi1-fi0; } } My problem is in the line: fi[0]+=fi1-fi0;. In the one-element vector fi[0] are wrong results. I read something about sum reduction, but I do not know how to do it during the calculation. Exist any simple solution of my problem?

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• Python and C++ Sockets converting packet data

  - by yeus

  First of all, to clarify my goal: There exist two programs written in C in our laboratory. I am working on a Proxy Server (bidirectional) for them (which will also mainpulate the data). And I want to write that proxy server in Python. It is important to know that I know close to nothing about these two programs, I only know the definition file of the packets. Now: assuming a packet definition in one of the C++ programs reads like this: unsigned char Packet[0x32]; // Packet[Length] int z=0; Packet[0]=0x00; // Spare Packet[1]=0x32; // Length Packet[2]=0x01; // Source Packet[3]=0x02; // Destination Packet[4]=0x01; // ID Packet[5]=0x00; // Spare for(z=0;z<=24;z+=8) { Packet[9-z/8]=((int)(720000+armcontrolpacket->dof0_rot*1000)/(int)pow((double)2,(double)z)); Packet[13-z/8]=((int)(720000+armcontrolpacket->dof0_speed*1000)/(int)pow((double)2,(double)z)); Packet[17-z/8]=((int)(720000+armcontrolpacket->dof1_rot*1000)/(int)pow((double)2,(double)z)); Packet[21-z/8]=((int)(720000+armcontrolpacket->dof1_speed*1000)/(int)pow((double)2,(double)z)); Packet[25-z/8]=((int)(720000+armcontrolpacket->dof2_rot*1000)/(int)pow((double)2,(double)z)); Packet[29-z/8]=((int)(720000+armcontrolpacket->dof2_speed*1000)/(int)pow((double)2,(double)z)); Packet[33-z/8]=((int)(720000+armcontrolpacket->dof3_rot*1000)/(int)pow((double)2,(double)z)); Packet[37-z/8]=((int)(720000+armcontrolpacket->dof3_speed*1000)/(int)pow((double)2,(double)z)); Packet[41-z/8]=((int)(720000+armcontrolpacket->dof4_rot*1000)/(int)pow((double)2,(double)z)); Packet[45-z/8]=((int)(720000+armcontrolpacket->dof4_speed*1000)/(int)pow((double)2,(double)z)); Packet[49-z/8]=((int)armcontrolpacket->timestamp/(int)pow(2.0,(double)z)); } if(SendPacket(sock,(char*)&Packet,sizeof(Packet))) return 1; return 0; What would be the easiest way to receive that data, convert it into a readable python format, manipulate them and send them forward to the receiver?

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• Instantiate type variable in Haskell

  - by danportin

  EDIT: Solved. I was unware that enabling a language extension in the source file did not enable the language extension in GHCi. The solution was to :set FlexibleContexts in GHCi. I recently discovered that type declarations in classes and instances in Haskell are Horn clauses. So I encoded the arithmetic operations from The Art of Prolog, Chapter 3, into Haskell. For instance: fac(0,s(0)). fac(s(N),F) :- fac(N,X), mult(s(N),X,F). class Fac x y | x -> y instance Fac Z (S Z) instance (Fac n x, Mult (S n) x f) => Fac (S n) f pow(s(X),0,0) :- nat(X). pow(0,s(X),s(0)) :- nat(X). pow(s(N),X,Y) :- pow(N,X,Z), mult(Z,X,Y). class Pow x y z | x y -> z instance (N n) => Pow (S n) Z Z instance (N n) => Pow Z (S n) (S Z) instance (Pow n x z, Mult z x y) => Pow (S n) x y In Prolog, values are insantiated for (logic) variable in a proof. However, I don't understand how to instantiate type variables in Haskell. That is, I don't understand what the Haskell equivalent of a Prolog query ?-f(X1,X2,...,Xn) is. I assume that :t undefined :: (f x1 x2 ... xn) => xi would cause Haskell to instantiate xi, but this gives a Non type-variable argument in the constraint error, even with FlexibleContexts enabled.

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• B-V to Kelvin formula

  - by PeanutPower

  Whilst looking for a "B-V color index to temperature conversion formula" I found this javascript: var C1 = 3.979145; var C2 = -0.654499; var C3 = 1.74069; var C4 = -4.608815; var C5 = 6.7926; var C6 = -5.39691; var C7 = 2.19297; var C8 = -.359496; bmv = parseFloat(BV); with (Math) { logt= C1 +C2*bmv +C3*pow(bmv,2) +C4*pow(bmv,3) +C5*pow(bmv,4) +C6*pow(bmv,5) +C7*pow(bmv,6) +C8*pow(bmv,7); t=pow(10,logt); } Which is supposed to convert B-V color index to temperature. Does anyone understand how this is working and if the output value is an approximation for temperature in celcius or kelvin? Is it something to do with products of logarithms?

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• How can I make an infinite cave using stage3d?

  - by ifree

  I want to make an infinite cave in my 3d game using flash stage3d. But I got no idea about how to build that cave. Can anyone can give me some solution or hint? update: I've tried agal fragment shader like squeae tunnel in shader toy code: var fragmentProgramCode:String = AGALUtils.build() .mov("ft0","v0") .div("ft1","ft0.xy","fc3.xy") .mul("ft2","fc6.x","ft1") .sub("ft3","ft2","fc5.x")//vec2 p = -1.0 + 2.0 * gl_FragCoord.xy / resolution.xy; .mul("ft1","ft3.x","ft3.x") .mul("ft2","ft3.y","ft3.y") .pow("ft4","ft1","fc6.z")//float r = pow( pow(p.x*p.x,16.0) + pow(p.y*p.y,16.0), 1.0/32.0 ); .pow("ft5","ft2","fc6.z") .add("ft1","ft4","ft5") .pow("ft4","ft1","fc6.w") .mov("ft5","fc5")//uv .sub("ft1","fc7.x","ft4") .add("ft5.x","fc7.x","ft1")//uv.x = .5*time + 0.5/r; .mov("ft6","fc0")//for atan .atan2("ft5.y","ft3.y","ft3.x",new <String>["fc7.y","fc5.x","fc7.z","fc7.w","fc8.x","fc8.y","fc8.z","fc8.w","fc9.x","fc9.y"],"ft6") .tex("ft0","ft5","fs0","repeat","linear","nomip")//tex .mul("ft1","ft4","ft4") .mul("ft2","ft1","ft4")//r*r*r .mul("ft1","ft0.xyz","ft2") .mov("ft0.w","fc5.x") .mov("oc","ft1").toString() it can only apply one material,but my project requires different types of material (like floor,ceilling). so ,I create a 3d model Is there anyway to make that 3d model render like "infinity cave"? use agal to make each side of cave's texture move? thanks for your help

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• What is more efficient? Using pow to square or just multiply it with itself?

  - by gnol

  What of these two methods is in C more efficient? And how about: pow(x,3) vs. x*x*x etc?

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• Why isnt int pow(int base, int exponent) in the standard C++ libraries?

  - by Dan O

  I feel like I must just be unable to find it. Is there any reason that the c++ pow function does not implement the "power" function for anything except floats and doubles? I know the implementation is trivial, I just feel like I'm doing work that should be in a standard library. A robust power function (ie handles overflow in some consistent, explicit way) is not fun to write.

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• power function in blackberry

  - by Rahul Varma

  Hi, I desperately need a solution for this... I m using Blackberry JDE 5.0. In documentation it shows net.rim.device.api.util.MathUtilities.pow ( double double ) is added form 4.6. But i couldnt get pow function when i use Math.pow(). I need to get this value...float fSecondFactor=(float) (Math.pow(1+fRatePerMonth,fNumberofMonths)-1); Can anyone tell me how to achieve this...Plzz....

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• Encode two integers into colour values and compare them in a HLSL shader

  - by Ben Slinger

  I am writing a 2D point and click adventure game in Monogame, and I'd like to be able to create an image mask for every room which defines which parts of the background a character can walk behind, and at which Y value a character needs to be at for the background to be drawn above the character. I haven't done any shader work before but after doing some reading I thought the following solution should work: Create a mask for the room with different walk behind areas painted in a colour that defines the baseline Y value (Walk Behind Mask) Render all objects to a RenderTarget2D (Base Texture) Render all objects to a different RenderTarget2D, but changing every pixel of each object to a colour that defines its Y value (Position Mask) Pass these two textures plus the image mask into the shader, and for each pixel compare the colour of the image mask to the colour of the Position Mask to the Walk Behind Mask - if the Position Mask pixel is larger (thus lower on the screen and closer to the camera) than the Walk Behind Mask, draw the pixel from the Base Texture, otherwise draw a transparent pixel (allowing the background to show through). I've got it mostly working, but I'm having trouble packing and unpacking the Y values into colours and retrieving them correctly in the shader. Here are some code examples of how I'm doing it so far: (When drawing to the Position Mask RenderTarget2D) Color posColor = new Color(((int)Position.Y >> 16) & 255, ((int)Position.Y >> 8) & 255, (int)Position.Y & 255); So as far as I can tell, this should be taking the first 3 bytes of the position integer and encoding them into a 4 byte colour (ignoring the alpha as the 4th byte). This seems to work fine, as when my character is at Y = 600, the resulting Color from this is: {[Color: R=0, G=2, B=88, A=255, PackedValue=4283957760]}. I then have an area in my Walk Behind Mask that I only want the character to be displayed behind if his Y value is lower than 655, so I've painted it with R=0, G=2, B=143, A=255. Now, I think I have the shader OK as well, here's what I have: sampler BaseTexture : register(s0); sampler MaskTexture : register(s1); sampler PositionTexture : register(s2); float4 mask( float2 coords : TEXCOORD0 ) : COLOR0 { float4 color = tex2D(BaseTexture, coords); float4 maskColor = tex2D(MaskTexture, coords); float4 positionColor = tex2D(PositionTexture, coords); float maskCompare = (maskColor.r * pow(2,24)) + (maskColor.g * pow(2,16)) + (maskColor.b * pow(2,8)); float positionCompare = (positionColor.r * pow(2,24)) + (positionColor.g * pow(2,16)) + (positionColor.b * pow(2,8)); return positionCompare < maskCompare ? float4(0,0,0,0) : color; } technique Technique1 { pass NoEffect { PixelShader = compile ps_3_0 mask(); } } This isn't working, however - currently all characters are displayed behind the walk behind area, regardless of their Y value. I tried printing out some debug info by grabbing the pixel from both the Position Mask and the Walk Under Mask under the current mouse position, and it seems like maybe the colours aren't being rendered to the Position Mask correctly? When calculating the colour in that code above I'm getting R=0, G=2, B=88, A=255, but when I mouseover my character I get R=0, G=0, B=30, A=255. Any ideas what I'm doing wrong? It seems like maybe I'm losing some information when rendering to the RenderTarget2D, but I'm now knowledgeable enough to figure out what's happening. Also, I should probably ask, is this an efficient way to do this? Will there be a performance impact? Edit: Whoops, turns out there was a bug that I'd introduced myself, I was drawing out the Position Mask with the position Color, left over from some early testing I was doing. So this solution is working perfectly, though I'm still interested in whether this is an efficient solution performance wise.

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