修改一个结构数组的值

Question

我目前正在尝试在C++中建模弹性碰撞。我有一个叫做粒子的结构，我用它来描述对象。

//Defining the particle structure 
struct particle { 
    double x; // Position 
    double p; // Momentum 
    double im; // Inverse Mass 
    double v; // Velocity 
    double T; // Kinetic Energy 
    double a; // Radius of a Particle 
}; 

我使用这些结构创建了一个数组。

particle AP[n+2]; 

我的程序的一部分需要使用时间变量和它们的速度增加每个结构的位置。

我使用这个代码

void LeapForward(struct particle AP[],double tfc,int n) 
{ 
    for(int cycle=0; cycle<n+2; cycle++) 
    { 
     double second; 
     second=AP[cycle].x+tfc*AP[cycle].v; 
     AP[cycle].x=second; 
    } 
} 

然而，看来正在修改的阵列心不是而是重新创建。我想知道是否有人知道这是为什么，如果是的话如何解决这个问题。

我的完整代码可以在下面找到。任何帮助将不胜感激。

//This program will be about particle collsions 
#include <iostream> 
#include <cmath> 
#include <fstream> 
using namespace std; 

//Function for collision of particles 
void Collision(struct particle AP[],int chosen); 
void Assignment(struct particle AP[],int n); 
void LeapForward(struct particle AP[], double tfc, int n); 
void TimeForCollision(struct particle AP[],double tfc, int chosen, int n); 

//Defining the particle structure 
struct particle { 
    double x; // Position 
    double p; // Momentum 
    double im; // Inverse Mass 
    double v; // Velocity 
    double T; // Kinetic Energy 
    double a; // Radius of a Particle 
}; 

//Variable Declaration 
int n; //Number of particles 
double tfc; //This is going to be the time to leap forward 
int chosen; //Which particles are to collide 

    //Main program 

    int main() 
    { 
    cout<<"How many particle do you want to model: "; 
    cin>>n; 
    particle AP[n+2];  

    //Assigning value to Particles 
    Assignment(AP,n);  

    cout<<endl<<"Two walls have been placed at x=0 and 20"<<endl; 
    //Assignment of Wall Data. 
    AP[0].im=0.000001,AP[n+1].im=0.000001; 
    AP[0].x=0,AP[n+1].v=0; 
    AP[0].x=0,AP[n+1].x=20; 
    AP[0].a=0,AP[n+1].a=0; 

    ofstream Port; 
    Port.open ("Positions.txt"); 
    for(int model=0; model<2;model++) 
     { 
    //Time for next collision 
    TimeForCollision(AP,tfc,chosen,n); 
    LeapForward(AP,tfc,n); 
    Collision(AP,chosen); 
    for(int plot=0; plot<n+2;plot++) 
     { 
     Port<<AP[plot].x<<"\t"; 
      } 
    Port<<endl; 
     } 
    } 


void Collision(struct particle AP[],int chosen) 
{ 
    double combinemass=1/AP[chosen].im+1/AP[chosen+1].im; 
    double result1=(AP[chosen].v*(1/(AP[chosen].im)-1/(AP[chosen+1].im))+(2*AP[chosen+1].v*1/(AP[chosen+1].im)))/combinemass; 
    double result2=(AP[chosen+1].v*(1/(AP[chosen+1].im)-1/(AP[chosen].im))+(2*AP[chosen].v*1/(AP[chosen].im)))/combinemass; 
    AP[chosen].v=result1; 
    AP[chosen+1].v=result2; 
} 

void Assignment(struct particle AP[],int n) 
{ 
for(int cycle=1;cycle<=n;cycle++) 
    { 
    cout<<"Data Input for particle number "<<cycle<<endl; 
    double mass; 
    cout<<"What is the position of the particle : "; 
    cin>>AP[cycle].x; 
    cout<<"What is the mass of the particle : "; 
    cin>>mass; 
    AP[cycle].im=1/mass; 
    cout<<"What is the velocity of the particle : "; 
    cin>>AP[cycle].v; 
    cout<<"What is the radius of the particle : "; 
    cin>>AP[cycle].a; 
    //Data calculation 
    AP[cycle].p=AP[cycle].v*mass; 
    AP[cycle].T=0.5*AP[cycle].v*mass*mass; 
    } 
} 

void LeapForward(struct particle AP[],double tfc,int n) 
{ 
    for(int cycle=0; cycle<n+2; cycle++) 
    { 
     double second; 
     second=AP[cycle].x+tfc*AP[cycle].v; 
     AP[cycle].x=second; 
    } 
} 

void TimeForCollision(struct particle AP[],double tfc, int chosen, int n) 
{ 
double balance=0; 
tfc=0; 
    for(int cycle=0;cycle<n+1;cycle++) 
    { 
     double placeh=abs(((AP[cycle+1].x-AP[cycle+1].a)-(AP[cycle].x+AP[cycle].a)))/(AP[cycle].v-AP[cycle+1].v); 
     if(placeh>tfc && placeh>0) 
     { 
     tfc=placeh; 
     chosen=cycle; 
     } 
    } 
    cout<<tfc<<endl; 
} 

Answer 1

这是因为当你打电话的功能，您将创建结构的新实例，并修改新创建的实例，这将在你的函数结束时被删除。

传递值作为这样一个参考：

void LeapForward(struct particle *AP, double tfc, int n); 

另外，在你LeapForward功能，您不需要创建临时的双变量。 您可以使用类似：

AP[cycle].x += tfc * AP[cycle].v; 

---

Live demo。