函数调用main和库中的不同行为

Function call different behavior in main and library

本文关键字:main 函数调用      更新时间:2023-10-16

我遇到了一件非常奇怪的事情,两天前我试图调试代码。我在Windows7 64位操作系统上运行代码。主要通过了解输入信号来计算数学模型,并将其应用于控制算法SOOP中。计算基本上是可以的,但在soop函数中,当我做同样的事情时,我得到了接近但不完全相同的结果。为什么?我也厌倦了float,我得到了同样的结果。

如果我在一个非主函数中计算,double是否会被取整

Main:

#include <unistd.h>
#include <iostream>
#include <windows.h>
#include <fstream>
#include <math.h>
#include "model.h"
#include "SOOP.h"
using namespace std;
int main( void )
{
    ofstream myfile;
    myfile.open ("modelTest.txt");
    stateE current_state; // = new stateE;
    initstateE(current_state);
    current_state.variables[0] = 0.0;
    current_state.variables[0] = -3.0*PI/2.0;
    cout <<  "program init" << endl;
   // ==============Model testing
    myfile << current_state.variables[0] <<", "<< current_state.variables[2] <<", "<<current_state.reward << ", "<< fmod((current_state.variables[0])+ PI, 2.0 * PI) -PI <<endl;
    double utmp = -10.0;
    for (int i=0;i<3;i++)
    {
        current_state.variables[0] = -3.0*PI/2.0;
        copyStateE(current_state, nextstateE(current_state, utmp));
        utmp = utmp+10.0;
        myfile << current_state.variables[0] <<", "<< current_state.variables[2] <<", "<<current_state.reward << ", "<< fmod((current_state.variables[0])+ PI, 2.0 * PI) -PI <<endl;
    }
     // ==============SOOP testing
     initstateE(current_state);
    current_state.variables[0] = -3.0*PI/2.0;
    double commnadSignal[initlen];
    for(int i=0; i<1; i++)
    {    
        memcpy(commnadSignal,soop(current_state),sizeof(double)*initlen);
        copyStateE(current_state,nextstateE(current_state, commnadSignal[0]));
    }
    myfile.close();
    return 0;
}

模型库:

stateE nextstateE(stateE s, double votlageSignal) {
    std::cout<< "in nextstateE"<< std::endl;
    stateE newstateE;// = (stateE*)malloc(sizeof(stateE));
    memcpy(newstateE.variables, s.variables, sizeof(double) * nbHiddenstateEVariables);
    newstateE.isTerminal = s.isTerminal;
    std::cout << " v "<< votlageSignal << " l "<< s.variables[0] << std::endl;
    double Vm = votlageSignal;
    double lambda = s.variables[0];
    double dLambda = s.variables[1];
    double theta = s.variables[2];
    double dTheta = s.variables[3];
    double thetaNext = theta + timeStep*dTheta;
    double dThetaNext = dTheta + timeStep*( (1/(aP*cP-pow(bP,2)*pow(cos(lambda),2))) *(-bP*cP*sin(lambda)*pow(dLambda,2)+bP*dP*sin(lambda)*cos(lambda)-cP*eP*dTheta+cP*fP*Vm) );
    double lambdaNext = lambda + timeStep*dLambda;
    double dLambdaNext = dLambda + timeStep*( (1/(aP*cP-pow(bP,2)*pow(cos(lambda),2))) *(aP*dP*sin(lambda)-pow(bP,2)*sin(lambda)*cos(lambda)*pow(dLambda,2)-bP*eP*cos(lambda)*dTheta+bP*fP*cos(lambda)*Vm) );
    //========== Normalization of lambda and theta
    if (lambdaNext>0)
            lambdaNext = fmod(lambdaNext+PI, 2.0*PI)-PI;
    else
            lambdaNext = fmod(lambdaNext-PI, 2.0*PI)+PI;
    if (thetaNext>0)
            thetaNext = fmod(thetaNext+PI, 2.0*PI)-PI;
    else
            thetaNext = fmod(thetaNext-PI, 2.0*PI)+PI;
    newstateE.variables[0] = lambdaNext;
    newstateE.variables[1] = dLambdaNext;
    newstateE.variables[2] = thetaNext;
    newstateE.variables[3] = dThetaNext;
    double newReward;
    if(s.isTerminal) {
        newReward = 0.0;
    } else {    
        newReward = 1 - (pow(lambdaNext,2.0)*qReward + pow(votlageSignal,2.0)*rReward)/(pow(PI,2.0)*qReward+ pow(maxVoltage,2.0)*rReward);
    }
    newstateE.reward=newReward;
    return newstateE;

}
void copyStateE(stateE& destination, stateE original)
{ 
    initstateE(destination);
    destination.variables[0] = original.variables[0];   // lambda
    destination.variables[1] = original.variables[1];   // dLambda
    destination.variables[2] = original.variables[2];   // theta
    destination.variables[3] = original.variables[3];   // dTheta
    destination.isTerminal = 0;
    destination.reward = original.reward;
}

控制器库使用模型库:

double * soop(stateE& s0)
{
    ofstream myfile;
    myfile.open ("soopTest.txt");
    bool stopFlag = false;
    unsigned int counter = 0;
    unsigned int currentBudget = 0;
    unsigned int listLenght = 0;
    unsigned int maxLengthSequence = 0;
    double tmpU = -10.0;
    for (int i=0;i<3;i++)
    {
        double sPtmp[initlen] = {0};
        unsigned int stmp[initlen] = {0};
        stateE newstateE[initlen];
        stateE tmpE;
        sPtmp[0] = ((double)i)/3.0;
        stmp[0] = 1;
        initstateE(tmpE);
        tmpE.variables[0]=-3.0*PI/2.0;
        copyStateE(tmpE, nextstateE(tmpE , tmpU ));
        tmpU = tmpU + 10.0;
            newstateE[0] = nextstateE(s0, unnormU(middleOfBox((double)i/3.0, stmp[0])) );

            myfile << "U norm "<< middleOfBox((double)i/3.0, stmp[0])<< " Unorm "<<unnormU(middleOfBox((double)i/3.0, stmp[0])) << endl;
            myfile <<"alp " << tmpE.variables[0] << " th " << tmpE.variables[2] <<  " R " << tmpE.reward << endl;
            myfile <<"alp " << newstateE[0].variables[0] << " th " << newstateE[0].variables[2] <<  " R " << newstateE[0] .reward << endl;
            myfile << endl;
    }
    myfile.close();
}

两个文本文件modelTest.txtsoop Test.txt应该显示相同的结果,因为我在main和soop函数中赞成相同的输入参数(s0和[-10,0,10])。但我在文本文件中得到了不同的结果。soop Test文件:

U norm 0.166667 Unorm -10
alp 1.5708 th 0 R 0.75
alp 1.5708 th 0 R 0.75
U norm 0.5 Unorm 0
alp 1.5708 th 0 R 0.75
alp 1.5708 th 0 R 0.75
U norm 0.833333 Unorm 10
alp 1.5708 th 0 R 0.75
alp 1.5708 th 0 R 0.75

modelTest文件:

alp 1.5708, th 0, R 0.75 (for s0 and -10)
alp 1.57084, th -0.000115852, R 0.749986 (for s0 and 0)
alp 1.57088, th -0.000230942, R 0.749972 (for s0 and 10)

我的错误是,在main中,我没有初始化current_state的所有状态(只有一个角度)。main循环中缺少initStateE(current_state)

下一个错误是,voltageSignal的奖励因子"reward=0",因此奖励不会改变控制值的函数。

最大的错误是假设系统的角状态参数在第一次模拟后会发生变化。我对f(Q(t),U(t))-一个二阶微分方程使用欧拉离散,所以控制信号只对第一次迭代中的一阶导数和第二次迭代后的位置有影响。

P_{k+1} = P_k + daltaT * Q_k 
Q_{k+1} = Q_k + deltaT * f(Q_k, U_k)

其中deltaT是采样时间。