Must be possible: 3d arrays

[Chaospling]

Hi there. I've read a lot archived threads now but I can't find the answer.
I thought it was straight forward to make a double[][][] but apparently it isn't as I've found that it's these buggers which makes my program crash...

I'm making a math program and I really need that third array. Can someone please tell me how it's possible?

My guess would be that I need a new header or that I have to call them something else than double or long double...

It would be highly appreciated - thanks for your time.

[Lv0f92yv]

It is possible, here is an example:

double a[5][6][5];

	for ( int i = 0; i < 5; i++ )
		for ( int j = 0; j < 6; j++ )
			for ( int k = 0; k < 5; k++ )
			{
				a[i][j][k] = j + k + i;
				cout << a[i][j][k] << "\n";
			}

Note that the condition is always less than the size allocated to the array.

Hope this helps.

[Chaospling]

Yes thank you. That's straight forward but my program crashes when having more than two arrays. I just wonder why it crashes.

[guestgulkan]

Look just post your code - or we'll just end up playing some stupid guessing game.

[Chaospling]

Okay thanks. This is some of the code, I don't think it's necessary to post it all.

void beregnH(double w[maxgrad], long double H[maxgrad][maxgrad][maxgrad], int kol, int mmax)
{
	double I[maxgrad][maxgrad];
	int i, j;
	for(i=0;i<mmax;i=i+1)
	{
		for(j=0;j<mmax;j=j+1)
		{
			I[i][j]=0;
			if(i==j)
			{
				I[i][i]=1;
			}
		}
	}
	for(j=0;j<mmax;j=j+1)
	{
		for(i=0;i<mmax;i=i+1)
		{
			H[i][j][kol]=I[i][j]-2*w[i]*w[j];
		}
	}
}

So "H" will consist of a "kol" number of matrices. But my C++ don't like more than two arrays.

[guestgulkan]

Chaospling wrote:

But my C++ don't like more than two arrays.

So what C++ (compiler) do you use?

[Chaospling]

An old one, C++ 6.0 but I though it still should be possible.

[Albatross]

Get... MinGW... or VS 2010 Express. That program... is from 1998.

-Alba... tross

[guestgulkan]

I'm sure it can handle more than 2 dimension arrays.
Post the whole code.

[Chaospling]

Hehe I know it's old. I just had problems with the newer ones and this is what's being used in my school. I don't think that I should post it all... It's quite long. But I can give you a little more.

void opgave4(double A[maxgrad][maxgrad], double b[maxgrad], int mmax, int nmax);
	long double H[maxgrad][maxgrad][maxgrad], R[maxgrad][maxgrad][maxgrad], Q[maxgrad][maxgrad][maxgrad];
	double l[maxgrad], x[maxgrad], reflx[maxgrad], w[maxgrad], Qt[maxgrad][maxgrad], Q1tb[maxgrad], R1Q1tb[maxgrad][maxgrad+1], xloes[maxgrad], I[maxgrad][maxgrad], tjekA[maxgrad][maxgrad];
	int kol;
	char svar;
	opgave4praesentation();
	do
	{
		cout<<"\n Oensker du at finde mindste kvadraters loesningen via grafisk forstaaelse, hvor du skal\n";
		cout<<" indtaste linien l, eller via metoden hvor ingen linie skal indtastes. Tast l hvis du vil\n";
		cout<<" bruge den grafiske forstaaelse: ";
		cin>>svar;
		if(svar=='l')
		{
			indtastl(A, l, mmax);
			for(kol=0;kol<nmax;kol=kol+1)
			{
				beregnx(A, R, x, kol, mmax);
				beregnreflx(x, l, reflx, mmax);
				beregnwl(x, reflx, w, mmax);
				beregnH(w, H, kol, mmax);
				beregnR(H, A, R, kol, mmax, nmax);
				beregnQ(H, Q, kol, mmax, nmax);
			}
		}
		else
		{
			for(kol=0;kol<nmax;kol=kol+1)
			{
				beregnx(A, R, x, kol, mmax);
				beregnw(x, w, kol, mmax);
				beregnH(w, H, kol, mmax);
				beregnR(H, A, R, kol, mmax, nmax);
				beregnQ(H, Q, kol, mmax, nmax);
			}
		}
		transponermatrix1(Q, Qt, mmax, nmax);
		beregnQ1tb(Qt, b, Q1tb, mmax, nmax);
		totalmatrix1(R, Q1tb, R1Q1tb, mmax, nmax);
	
		cout<<"\n\n Vi har nu totalmatricen:\n\n";
		udskrivtotalmatrix(R1Q1tb, nmax);
	
		backwardssubstitution(R1Q1tb, xloes, nmax);
	
		cout<<"\n\n Her er mindste kvadraters loesningen:\n";
		udskrivloesning(xloes, nmax);
		tjekloesning1(Qt, Q, I, mmax, nmax);

		cout<<"\n Da Q er en ortonormal matrix burde Q transponeret ganget med Q meget gerne give enhedsmatricen.\n";
		cout<<" Her er hvad Qt*Q bliver:\n\n";
		udskrivmatrix2(I, nmax);
		tjekloesningA1(Q, R, tjekA, mmax, nmax);

		cout<<"\n Vi har jo at A=Q*R. Saa her har du foerst din indtastet matrix A og efterfoelgende matrix A beregnet\n";
		cout<<" via Q*R som altsaa meget gerne skulle ligne den foerste.\n\n";
		udskrivmatrix(A, mmax, nmax);
		cout<<"\n\n";
		udskrivmatrix(tjekA, mmax, nmax);

		cout<<"\n Oensker du at finde mindste kvadraters loesningen igen ved brug af Householder reflections? ";
		cin>>svar;
	}
	while(svar=='j');
}


void opgave4praesentation()
{

}


void indtastl(double A[maxgrad][maxgrad], double v[maxgrad], int mmax)
{
	double temp, temp1;
	int i;
	char svar;
	cout<<"\n Indtast elementerne for den faste linie/vektor l, som der skal reflekteres om.\n";
	do
	{
		do
		{
			for(i=0;i<mmax;i=i+1)
			{
				cout<<"\n Indtast element nr."<<i+1<<" i vektoren her: ";
				cin>>v[i];
			}
			temp=A[0][0]/v[0];
			i=1;
			do
			{
				temp1=A[i][0]/v[i];
				i=i+1;
			}
			while(temp==temp1&&i<mmax);
			if(temp==temp1)
			{
				cout<<"\n Den indtastet vektor har samme retning som den foerste kolonne i matrix A, saa indtast\n";
				cout<<" en ny vektor\n.";
			}
		}
		while(temp==temp1);
		cout<<"\n Dette er den indtastet linie der skal reflekteres om:\n\n";
		udskrivvektor(v, mmax);
		cout<<"\n Vil du indtaste elementerne om? ";
		cin>>svar;
	}
	while(svar=='j');
}


void beregnreflx(double x[maxgrad], double l[maxgrad], double reflx[maxgrad], int mmax)
{
	double temp, temp1, temp2, projx[maxgrad];
	int i;
	temp1=0;
	temp2=0;
	for(i=0;i<mmax;i=i+1)
	{
		temp1=x[i]*l[i]+temp1;
		temp2=pow(l[i],2)+temp2;
	}
	temp=temp1/temp2;
	for(i=0;i<mmax;i=i+1)
	{
		projx[i]=temp*l[i];
		reflx[i]=2*projx[i]-x[i];
	}
}


void beregnwl(double x[maxgrad], double reflx[maxgrad], double w[maxgrad], int mmax)
{
	double temp, temp1, temp2;
	int i;
	temp=0;
	for(i=0;i<mmax;i=i+1)
	{
		temp2=x[i]-reflx[i];
		temp=pow(temp2,2)+temp;
	}
	temp1=1/sqrt(temp);
	for(i=0;i<mmax;i=i+1)
	{
		w[i]=temp1*(x[i]-reflx[i]);
	}
}


void beregnH(double w[maxgrad], long double H[maxgrad][maxgrad][maxgrad], int kol, int mmax)
{
	double I[maxgrad][maxgrad];
	int i, j;
	for(i=0;i<mmax;i=i+1)
	{
		for(j=0;j<mmax;j=j+1)
		{
			I[i][j]=0;
			if(i==j)
			{
				I[i][i]=1;
			}
		}
	}
	for(j=0;j<mmax;j=j+1)
	{
		for(i=0;i<mmax;i=i+1)
		{
			H[i][j][kol]=I[i][j]-2*w[i]*w[j];
		}
	}
}


void beregnR(long double H[maxgrad][maxgrad][maxgrad], double A[maxgrad][maxgrad], long double R[maxgrad][maxgrad][maxgrad], int kol, int mmax, int nmax)
{
	int i, j, k;
	for(i=0;i<mmax;i=i+1)
	{
		for(j=0;j<nmax;j=j+1)
		{
			R[i][j][kol]=0;
		}
	}
	if(kol==0)
	{
		for(k=0;k<nmax;k=k+1)
		{
			for(i=0;i<mmax;i=i+1)
			{
				for(j=0;j<mmax;j=j+1)
				{
					R[i][k][kol]=H[i][j][kol]*A[j][k]+R[i][k][kol];
				}
			}
		}
	}
	else
	{
		for(k=0;k<nmax;k=k+1)
		{
			for(i=0;i<mmax;i=i+1)
			{
				for(j=0;j<mmax;j=j+1)
				{
					R[i][k][kol]=H[i][j][kol]*R[j][k][kol-1]+R[i][k][kol];
				}
			}
		}
	}
}


void beregnQ(long double H[maxgrad][maxgrad][maxgrad], long double Q[maxgrad][maxgrad][maxgrad], int kol, int mmax, int nmax)
{
	int i, j, k;
	for(i=0;i<mmax;i=i+1)
	{
		for(j=0;j<mmax;j=j+1)
		{
			Q[i][j][kol]=0;
		}
	}
	if(kol==0)
	{
		for(i=0;i<mmax;i=i+1)
		{
			for(j=0;j<mmax;j=j+1)
			{
				Q[i][j][kol]=H[i][j][kol];
			}
		}
	}
	else
	{
		for(k=0;k<mmax;k=k+1)
		{
			for(i=0;i<mmax;i=i+1)
			{
				for(j=0;j<mmax;j=j+1)
				{
					Q[i][k][kol]=Q[i][j][kol-1]*H[j][k][kol]+Q[i][k][kol];
				}
			}
		}
	}
}


void beregnx(double A[maxgrad][maxgrad], long double R[maxgrad][maxgrad][maxgrad], double x[maxgrad], int kol, int mmax)
{
	int i;
	if(kol==0)
	{
		for(i=0;i<mmax;i=i+1)
		{
			x[i]=A[i][0];
		}
	}
	else
	{
		for(i=0;i<mmax;i=i+1)
		{
				x[i]=R[i][kol][kol-1];
		}
	}
}


void beregnw(double x[maxgrad], double w[maxgrad], int kol, int mmax)
{
	double yk, xk, sk, xkyk;
	int i;
	yk=0;
	for(i=kol;i<mmax;i=i+1)
	{
		yk=pow(x[i],2)+yk;
	}
	if(x[kol]<0)
	{
		yk=sqrt(yk);
		sk=yk;
		xk=-x[kol];
	}
	if(x[kol]>0)
	{
		yk=-sqrt(yk);
		sk=-yk;
		xk=x[kol];
	}
	xkyk=x[kol]-yk;
	w[kol]=xkyk;
	if(kol>0)
	{
		for(i=0;i<kol;i=i+1)
		{
			w[i]=0;
		}
	}
	if(kol<mmax)
	{
		for(i=kol+1;i<mmax;i=i+1)
		{
			w[i]=x[i];
		}
	}
	for(i=0;i<mmax;i=i+1)
	{
		w[i]=w[i]/sqrt(2*sk*(sk+xk));
	}
}


void transponermatrix1(long double A[maxgrad][maxgrad][maxgrad], double At[maxgrad][maxgrad], int mmax, int nmax)
{
	int i, j;
	for(i=0;i<mmax;i=i+1)
	{
		for(j=0;j<nmax;j=j+1)
		{
			At[j][i]=A[i][j][nmax-1];
		}
	}
}


void beregnQ1tb(double At[maxgrad][maxgrad], double b[maxgrad], double Atb[maxgrad], int mmax, int nmax)
{
	int i, j;
	for(i=0;i<nmax;i=i+1)
	{
		Atb[i]=0;
	}
	for(i=0;i<nmax;i=i+1)
	{
		for(j=0;j<mmax;j=j+1)
		{
			Atb[i]=At[i][j]*b[j]+Atb[i];
		}
	}
}


void totalmatrix1(long double AtA[maxgrad][maxgrad][maxgrad], double Atb[maxgrad], double R1Q1tb[maxgrad][maxgrad+1], int mmax, int nmax)
{
	int i, j;
	for(i=0;i<nmax;i=i+1)
	{
		for(j=0;j<nmax;j=j+1)
		{
			R1Q1tb[i][j]=AtA[i][j][nmax-1];
		}
	}
	for(i=0;i<nmax;i=i+1)
	{
		R1Q1tb[i][nmax]=Atb[i];
	}
}


void tjekloesning1(double Qt[maxgrad][maxgrad], long double Q[maxgrad][maxgrad][maxgrad], double I[maxgrad][maxgrad], int mmax, int nmax)
{
	double temp;
	int i, j, n;
	for(i=0;i<nmax;i=i+1)
	{
		for(j=0;j<nmax;j=j+1)
		{
			I[i][j]=0;
		}
	}
	for(n=0;n<nmax;n=n+1)
	{
		for(j=0;j<nmax;j=j+1)
		{
			temp=0;
			for(i=0;i<mmax;i=i+1)
			{
				temp=Qt[j][i]*Q[i][n][nmax-1]+temp;
			}
			I[i][n]=temp;
		}
	}
}


void tjekloesningA1(long double Q[maxgrad][maxgrad][maxgrad], long double R[maxgrad][maxgrad][maxgrad], double tjekA[maxgrad][maxgrad], int mmax, int nmax)
{
	int i, j, n;
	for(i=0;i<mmax;i=i+1)
	{
		for(j=0;j<nmax;j=j+1)
		{
			tjekA[i][j]=0;
		}
	}
	for(n=0;n<nmax;n=n+1)
	{
		for(i=0;i<mmax;i=i+1)
		{
			for(j=0;j<nmax;j=j+1)
			{
				tjekA[i][n]=Q[i][j][nmax-1]*R[j][n][nmax-1]+tjekA[i][n];
			}
		}
	}
}