# 这段程序代码怎么改？？？？？？

``````#include "iostream"
using namespace std;
class Gelr
{
public:
void gelr1(double t,double y[] ,int n,double h,int k,double z[][11]);
void gelrf(double t,double y[],int n,double d[]);
//void display();
protected:
double t;
double y[3];
int n;
double h;
int k;

double z[3][11];
double d[3];
};
void Gelr::gelr1(double t,double y[],int n,double h,int k, double x[])
{
int i,j;
double x,*d;
d=static_cast(malloc(n*sizeof(double)));
for (i=0; i<=n-1; i++) z[i*k]=y[i];
for (j=1; j<=k-1; j++)
{ x=t+(j-1)*h;
gelrf(x,y,n,d);
for (i=0; i<=n-1; i++)
y[i]=z[i*k+j-1]+h*d[i];
x=t+j*h;
gelrf(x,y,n,d);
for (i=0; i<=n-1; i++)
d[i]=z[i*k+j-1]+h*d[i];
for (i=0; i<=n-1; i++)
{ y[i]=(y[i]+d[i])/2.0;
z[i*k+j]=y[i];
}
}
free(d); return;
}
void Gelr::gelrf(double t,double y[],int n,double d[])
{
t=t;
n=n;
d[0]=y[1];
d[1]=-y[0];
d[2]=-y[2];
return;
}
int main()
{
int i,j;
double y[3],z[3][11],t,h,x;
y[0]=-1.0; y[1]=0.0; y[2]=1.0;
t=0.0; h=0.01;
Gelr g;
g.gelr1(t,y,3,h,11,z);
cout<<"\n"<<endl;
for (i=0; i<=10; i++)
{ x=i*h;
cout<<"t:"<<x<<endl;
for (j=0; j<=2; j++)
cout<<"y("<<j<<")=%"<<z[j][i]<<endl;
cout<<"\n"<<endl;
}
cout<<"\n"<<endl;
return 0;
}``````

--------------------Configuration: sdt - Win32 Debug--------------------
Compiling...
sdt.cpp
D:防灾C++程序设计电子教案微分方程stdsdt.cpp(19) : error C2511: 'gelr1' : overloaded member function 'void (double,double [],int,double,int,double [])' not found in 'Gelr'
D:防灾C++程序设计电子教案微分方程stdsdt.cpp(4) : see declaration of 'Gelr'

sdt.exe - 1 error(s), 0 warning(s)

a123456678 2016-03-04 18:28:11 1880 0
1 条回答

• 在这段代码中，有以下问题：
1.gelr1函数前没有类的作用域运算符，那么它就不是类的成员函数，g.gelr1(t,y,3,h,11,z)这样的调用要出错；
2.gelr1中数组z的类型不匹配；
3.static_cast的用法错了；
你把代码改成这样：

``````void Gelr::gelr1(double t,double y[],int n,double h,int k, double z[][11])
{
int i,j;
double x,*d;
d=static_cast(malloc(n*sizeof(double)));
for (i=0; i<=n-1; i++)
z[i][0]=y[i];
for (j=1; j<=k-1; j++)
{ x=t+(j-1)*h;
gelrf(x,y,n,d);
for (i=0; i<=n-1; i++)
y[i]=z[i][j-1]+h*d[i];
x=t+j*h;
gelrf(x,y,n,d);
for (i=0; i<=n-1; i++)
d[i]=z[i][j-1]+h*d[i];
z[i][j]=y[i];
for (i=0; i<=n-1; i++)
{ y[i]=(y[i]+d[i])/2.0;
}
}
free(d);
return;
}``````
2019-07-17 18:52:43
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