为什么我的c++;代码将进入无限循环? 编写了一个LBM的C++代码求解扩散方程,并在Ubuntu中用G++编译。它编译得很好,但是当我尝试执行程序时,它并没有结束。我以为程序有一个无限循环,但我不能解决在哪里?提前谢谢你们
这是我的密码:为什么我的c++;代码将进入无限循环? 编写了一个LBM的C++代码求解扩散方程,并在Ubuntu中用G++编译。它编译得很好,但是当我尝试执行程序时,它并没有结束。我以为程序有一个无限循环,但我不能解决在哪里?提前谢谢你们,c++,g++,infinite-loop,C++,G++,Infinite Loop,这是我的密码: #include <iostream> #include <fstream> using namespace std; int main(){ const int Lx = 100; const int Ly = 100; int sMod = 8; //speed model float f[sMod][Lx+1][Ly+1]; float rho[Lx+1][Ly+1]; float feq, su
#include <iostream>
#include <fstream>
using namespace std;
int main(){
const int Lx = 100;
const int Ly = 100;
int sMod = 8; //speed model
float f[sMod][Lx+1][Ly+1];
float rho[Lx+1][Ly+1];
float feq, sum;
float x[Lx+1];
float y[Ly+1];
float csq, alpha, omega,Tw;
float w[9] = {4./9., 1./9., 1./9., 1./9., 1./9., 1./36., 1./36., 1./36., 1./36.};
int i, j, dt, dx, k, dy;
FILE * mFile;
dt = 1.0;
dx = 1.0;
dy = dx;
x[0] = 0.0;
for(i = 1; i < Lx; i++){
x[i] = x[i-1] + dx;
}
y[0] = 0.0;
for(j = 1; j < Ly; j++){
y[j] = y[j-1] + dy;
}
Tw = 1.0;
csq = (dx * dx)/(dt * dt);
alpha = 0.25;
omega = 1.0/((3.*alpha/(dt*csq))+0.5);
cout << "csq: " << csq << ", omega: " << omega << endl;
int mstep = 2;
for(j = 0; j <= Ly; j++)
{
for(i = 0; i <= Lx; i++)
{
rho[i][j] = 0.0;
//cout << "1: " << rho[0][Ly/2] << endl;
}
}
//cout << "5: " << rho[0][Ly/2] << endl;
for(j = 0; j <= Ly; j++)
{
//cout << "2: " << rho[0][Ly/2] << endl;
for(i = 0; i <= Lx; i++)
{
//cout << "3: " << rho[0][Ly/2] << endl;
for(k = 0; k <= sMod; k++)
{
//cout << "4: " << rho[0][Ly/2] << endl;
f[k][i][j] = w[k] * rho[i][j];
if(i == 0){
f[k][i][j] = w[k] * Tw;
}
}
}
}
cout << "2: " << rho[0][Ly/2] << endl;
for(k = 1; k <= mstep; k++)
{
for(j = 0; j <= Ly; j++)
{
for(i = 0; i <= Lx; i++)
{
sum = 0.0;
for(k = 0;k <= sMod; k++)
{
sum = sum + f[k][i][j];
}
rho[i][j] = sum;
}
}
cout << rho[0][Ly/2] << endl;
for(j = 0; j <= Ly; j++)
{
for(i = 0; i <= Lx; i++)
{
for(k = 0;k <= sMod; k++)
{
feq = w[k] * rho[i][j];
f[k][i][j] = omega * feq + (1. - omega) * f[k][i][j];
}
}
}
/*-------
streaming
--------*/
for(j = Ly; i >= 0; j--)
{
for(i = 0; i <= Lx; i++)
{
f[2][i][j] = f[2][i][j-1];
f[6][i][j] = f[6][i+1][j-1];
}
}
for(j = Ly; i >= 0; j--)
{
for(i = Lx; i >= 0; i--)
{
f[1][i][j] = f[1][i-1][j]; //right to left
f[5][i][j] = f[5][i-1][j-1];
}
}
for(j = 0; i <= Ly; j--)
{
for(i = Lx; i >= 0; i--)
{
f[4][i][j] = f[4][i][j+1];
f[8][i][j] = f[8][i-1][j+1];
}
}
for(j = 0; j <= Ly; j++)
{
for(i = 0; i <= Lx; i++)
{
f[3][i][j] = f[3][i+1][j];
f[7][i][j] = f[7][i+1][j+1];
}
}
//boundary conditions
for(j = 0; j <= Ly; j++)
{
f[1][0][j] = w[1]*Tw + w[3]*Tw - f[3][0][j];
f[5][0][j] = w[5]*Tw + w[7]*Tw - f[7][0][j];
f[8][0][j] = w[8]*Tw + w[6]*Tw - f[6][0][j];
f[3][Lx][j] = -f[1][Lx][j];
f[6][Lx][j] = -f[8][Lx][j];
f[7][Lx][j] = -f[5][Lx][j];
}
for(i = 0; i <= Lx; i++)
{
f[4][i][Ly] = -f[2][i][Ly];
f[7][i][Ly] = -f[5][i][Ly];
f[8][i][Ly] = -f[6][i][Ly];
f[1][i][0] = f[1][i][1];
f[2][i][0] = f[2][i][1];
f[3][i][0] = f[3][i][1];
f[4][i][0] = f[4][i][1];
f[5][i][0] = f[5][i][1];
f[6][i][0] = f[6][i][1];
f[7][i][0] = f[7][i][1];
f[8][i][0] = f[8][i][1];
}
}
for(j = 0; j <= Ly; j++)
{
for(i = 0; i <= Lx; i++)
{
sum = 0.0;
for(k = 0;k <= sMod; k++)
{
sum = sum + f[k][i][j];
}
rho[i][j] = sum;
}
}
cout << rho[0][Ly/2] << endl;
mFile = fopen("lbmdiffusiond2q9.csv","w");
fprintf(mFile,"\t\t----TITLE= D2Q9 RESULTS----\n");
fprintf(mFile," VARIABLES = RHO\n");
//myfile << "ZONE " << "I=" << Lx+1 << " J=" << Ly+1 << " F=POINT" << endl;
for(i = 0; i <= Lx; i++)
{
for(j = 0; j <= Ly; j++)
{
//myfile << (dx/Lx) * i << " " << (dy/Ly) * j << " " << rho[i][j] << endl;
fprintf(mFile,"%.6f ",rho[i][j]);
}
fprintf(mFile,"\n");
}
fclose(mFile);
return 0;
}
#包括
#包括
使用名称空间std;
int main(){
常数int Lx=100;
常数int-Ly=100;
int sMod=8;//速度模型
浮动f[sMod][Lx+1][Ly+1];
浮动rho[Lx+1][Ly+1];
浮动feq,总和;
浮动x[Lx+1];
浮动y[Ly+1];
浮动csq,α,ω,Tw;
浮动w[9]={4./9,1./9,1./9,1./9,1./9,1./9,1./36,1./36,1./36,1./36.};
int i,j,dt,dx,k,dy;
文件*mFile;
dt=1.0;
dx=1.0;
dy=dx;
x[0]=0.0;
对于(i=1;i=0;j--)的cout
{
对于(i=0;i这一点似乎是您的问题:
for(j = 0; i <= Ly; j--)
{
for(i = Lx; i >= 0; i--)
{
f[4][i][j] = f[4][i][j+1];
f[8][i][j] = f[8][i-1][j+1];
}
}
(j=0;i=0;i--)的
{
f[4][i][j]=f[4][i][j+1];
f[8][i][j]=f[8][i-1][j+1];
}
}
外部循环从0开始,减少计数器,循环条件不检查计数器。事实上,如果在循环之前为真,则循环条件保证永远为真,因为内部循环将以i==0
结束。在这种情况下,我通常会删除所有循环,然后检查每个循环现在不是看哪一个的时候是一个问题。谢谢。我更正了代码并重新编译了它,但它仍然是一个无限循环。还有一个地方,您使用k
作为外部和内部循环索引,您也解决了吗?使用更有意义的变量名,而不是I、j、k等,可以帮助您避免这些类型的错误。此外,它还必须使用调试器甚至一些print语句来识别程序被卡住的位置是非常简单的。
for(j = 0; i <= Ly; j--)
{
for(i = Lx; i >= 0; i--)
{
f[4][i][j] = f[4][i][j+1];
f[8][i][j] = f[8][i-1][j+1];
}
}