CODEWITHSUNDEEP
c++operator-overloadingpass-by-reference
jojo123456
jojo123456

Reputation: 341

cpp - large array pass by reference

I have a very large array which has around 10000000 double entries (in the code below this number was set to NX=1000 but actually it will be 10000000 in realistic situations). I want to do some calculations with this array from which I get a new array. Then I do again some calculations with this new array and so on. I am looking for the most efficient way to do these manipulations with respect to RAM (I believe that this is the limiting factor here, since my arras are so large). I know that because of the large arrays I should try to pass the array as much as possible per reference. However I have problems with passing the arrays to other functions. Here is the code:

#include <iostream>
#include <array>


#define NX 1000

////////////////////////
//Function declaration//
////////////////////////
void time_propagation(std::array<double,NX>& X1,std::array<double,NX>& X2, double dt); 

std::array<double,NX> F(std::array<double,NX>& X);

///////////////////////
//operator overloading/
///////////////////////

// double* Array
std::array<double,NX> operator*(double d,std::array<double,NX> X){
  std::array<double,NX> res={};
  for(int i=0;i<NX;i++){
    res[i]=X[i]*d;
  } 
  return res;
}

//Array + Array
std::array<double,NX> operator+(std::array<double,NX> X1,std::array<double,NX> X2){
  std::array<double,NX> res={};
  for(int i=0;i<NX;i++){
    res[i]=X1[i]+X2[i];
  }
  return res;
}

/////////////
//main()/////
/////////////

int main () {
  std::array<double,NX> X0={};  
  std::array<double,NX> X1 = {};
  time_propagation(X0,X1,0.1);
  //Now I will write X0 to a file. Afterwards X0 is not needed anymore. Only X1. Do I have to manually delete X0?
  std::array<double,NX> X2 = {};
  time_propagation(X1,X2,0.1);
  //Now I will write X1 to a file. Afterwards X1 is not needed anymore. Only X2. Do I have to manually delete X1?
  //... I have to do this step very often
 
 
  return 0;
}

///////////////////////
//Function definitons//
///////////////////////

std::array<double,NX> F(std::array<double,NX>& X){
  std::array<double,NX> R={};
  double a=X[NX-2];
  double b=X[NX-1];
  R[NX-1]=-a;
  R[NX-1]=3.0;
  
  return R;
}

void time_propagation(std::array<double,NX>& X1,std::array<double,NX>& X2,double dt){
  
  std::array<double,NX> K1=F(X1);
  std::array<double,NX> K2=F(X1+dt/2.0*K1);
  std::array<double,NX> K3=F(X1+dt/2.0*K2);
  std::array<double,NX> K4=F(X1+dt*K3);      
  X2=X1+1.0/6.0*dt*(K1+2.0*K2+2.0*K3+K4); 
}

I know that my error is in the function "time_propagation". In the fifth last line I call the function "F" with the argument "X1+dt/2.0*K1". "X1" was passed by reference and "K1" was defined before. Do I first have to create a copy and the call the function on that copy? Something like that:

 void time_propagation(std::array<double,NX>& X1,std::array<double,NX>& X2,double dt){
      
  std::array<double,NX> argument={};
  for(int i=0;i<NX;i++){argument[i]=X1[i];}      

  std::array<double,NX> K1=F(argument);
  for(int i=0;i<NX;i++){argument[i]=X1[i]+dt/2.0*K1[i];}   
  std::array<double,NX> K2=F(argument);
  for(int i=0;i<NX;i++){argument[i]=X1[i]+dt/2.0*K2[i];}   
  std::array<double,NX> K3=F(argument);
  for(int i=0;i<NX;i++){argument[i]=X1[i]+dt/2.0*K3[i];}   
  std::array<double,NX> K4=F(argument);   
  for(int i=0;i<NX;i++){X2[i]=X1[i]+1.0/6.0*dt*(K1[i]+2.0*K2[i]+2.0*K3[i]+K4[i]); }      
}

If I use this new "time_propagation" function then I don't get any errors anymore, but my feeling is that this workaround is very bad and there is a nicer way of doing this? Second and most importantly even with this new version of "time_propagation()", I get a "Segmentation fault (core dumped)" if I increase NX to 10000000. I guess this happens because the code is very badly written? In general my laptop should be able to store 10000000 doubles in the RAM? I have 16 GB of RAM... and 10000000 doubles require only 0.2 GB of RAM!?

Upvotes: 0

Views: 254

Answers (1)

prehistoricpenguin
prehistoricpenguin

Reputation: 6326

According to your comment, you are very close to the answer. One thing you missed is that besides array is stack-based and vector is heap-based there is one more difference between array and vector: the vector needs to call its constructor(or resize) to enlarge size. So after switching to vector, we need a slight modify.

  1. How to reduce memory usage?

Change the unnecessary copy arguments to reference.

  1. How to manually recycle a vector?

Use a lambda to call swap or clear and shrink_to_fit

Bonus: To remove the unnecessary temporary object in operator * and operator +, see Expression_templates

A fixed code might like this:

#include <array>
#include <iostream>
#include <vector>

#define NX 1000

////////////////////////
// Function declaration//
////////////////////////
void time_propagation(std::vector<double>& X1, std::vector<double>& X2,
                      double dt);

std::vector<double> F(const std::vector<double>& X);

///////////////////////
// operator overloading/
///////////////////////

// double* Array
std::vector<double> operator*(double d, const std::vector<double>& X) {
  std::vector<double> res(NX);
  for (int i = 0; i < NX; i++) {
    res[i] = X[i] * d;
  }
  return res;
}

// Array + Array
std::vector<double> operator+(const std::vector<double>& X1,
                              const std::vector<double>& X2) {
  std::vector<double> res(NX);
  for (int i = 0; i < NX; i++) {
    res[i] = X1[i] + X2[i];
  }
  return res;
}

/////////////
// main()/////
/////////////

int main() {
  auto release_vec = [](auto& vec) {
    vec.clear();
    vec.shrink_to_fit();
  };
  std::vector<double> X0(NX);
  std::vector<double> X1(NX);
  time_propagation(X0, X1, 0.1);
  // Now I will write X0 to a file. Afterwards X0 is not needed anymore. Only
  // X1. Do I have to manually delete X0?
  release_vec(X0);
  std::vector<double> X2(NX);
  time_propagation(X1, X2, 0.1);
  // Now I will write X1 to a file. Afterwards X1 is not needed anymore. Only
  // X2. Do I have to manually delete X1?
  //... I have to do this step very often
  release_vec(X1);

  return 0;
}

///////////////////////
// Function definitons//
///////////////////////

std::vector<double> F(const std::vector<double>& X) {
  std::vector<double> R(NX);
  double a = X[NX - 2];
  double b = X[NX - 1];
  R[NX - 1] = -a;
  R[NX - 1] = 3.0;

  return R;
}

void time_propagation(std::vector<double>& X1, std::vector<double>& X2,
                      double dt) {
  std::vector<double> K1 = F(X1);
  std::vector<double> K2 = F(X1 + dt / 2.0 * K1);
  std::vector<double> K3 = F(X1 + dt / 2.0 * K2);
  std::vector<double> K4 = F(X1 + dt * K3);
  X2 = X1 + 1.0 / 6.0 * dt * (K1 + 2.0 * K2 + 2.0 * K3 + K4);
}

Upvotes: 1

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