Segment fail on fortran 90 code

[Andrewgaven]

Hello all, i'm a newbie to fortran 90, i believe that there a mistake with my variable er3 declaration but i can't find it. my code is:
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SUBROUTINE BACKWARD(E,Y,p,L,k,pqr,O5,er5,er4,er3)
IMPLICIT NONE
INTEGER , INTENT(IN) :: p,k,L
INTEGER :: i,j,jo
REAL , INTENT(IN) :: E(p,k),Y(p,1),pqr(L**k,k+1),O5(1,p)
REAL , INTENT(OUT) :: er5(1,p),er4(L**k,p),er3(L**k,p)
REAL :: T(1,p),U(1,k+1),pal(L**k,2),pala(L**k,p)
REAL(kind=4) :: err
T=transpose(Y);
er5=-2*(T-O5);
U(1,k+1)=1.;
do j=1,p
er4,j)=(/ (er5(1,j), i=1,L**k) /)
do i=1,k
U(1,i)=E(j,i);
enddo
pala,j)=(/ (matmul(pqr,transpose(U))) /)
enddo
do i=1,L**k
er3(i,=(/ (er4(i,j)*pala(i,j), j=1,p)/)
enddo
do i=1,3
write(*,*) er3(i,
enddo
end SUBROUTINE
------------------------------
went executed this code shows the matrix er3 followed by
"fallo de segmentacion" (Spanish for "Segment fail").

Does anyone see anything wrong?
thanks in advance.

 

[xwb]

What sort of value does L**k end up as?

In the call to transpose y, how does it know that Y is of size (p,1)

 

[Andrewgaven]

Thank for the reply
these variables are defined in the main program but are related to the others input.
[p,k]=size(E); --> this is scilab code
"E" is the data matrix entry, "k" is the number column, and p the number or rows
"Y" is the data matrix output, it is only a single column and p rows.
"L" is an integer number it indicates the number of membership function because this is intended to be use in a fuzzy inference system.
The dimension of the other variables is always dependent o L,p and k.

this code works as my main program and defines all the variables need it.
----------------------------------------------------
integer k,L,p
real E(p,k),Y(p,1),abc(3*L*k,1),pqr(L,k+1),O5(1,p),O4(L**k,p),O3(L**k,p),O2(L**k,p),O1(L**k,p),er5(1,p),er4(L**k,p)
p=10;L=3;k=1;

E(1,1)=10700;E(2,1)=10700;E(3,1)=10700;E(4,1)=10600;E(5,1)=10600;E(6,1)=10600;E(7,1)=10600;E(8,1)=10600;E(9,1)=10500;E(10,1)=10600;

Y(1,1)=12536;Y(2,1)=12536;Y(3,1)=12532;Y(4,1)=12532;Y(5,1)=12532;Y(6,1)=12532;Y(7,1)=12528;Y(8,1)=12528;Y(9,1)=12528;Y(10,1)=12524;

abc(1,1)=50.;abc(2,1)=2.;abc(3,1)=10500.;abc(4,1)=50.;abc(5,1)=2.;abc(6,1)=10600.;abc(7,1)=50.;abc(8,1)=2.;abc(9,1)=10700.;

pqr(1,1)=1.1938868;pqr(1,2)=0.0001136;pqr(2,1)=1.1819761;pqr(2,2)=0.0001115;pqr(3,1)=1.1707586;pqr(3,2)=0.0001095;

O5(1,1)=12534.667;O5(1,2)=12534.667;O5(1,3)=12534.667;O5(1,4)=12529.333;O5(1,5)=12529.333;O5(1,6)=12529.333;O5(1,7)=12529.333;
O5(1,8)=12529.333;O5(1,9)=12528.;O5(1,10)=12529.333;

O4(1,1)=46.77321;O4(1,2)=46.77321;O4(1,3)=46.77321;O4(1,4)=666.06318;O4(1,5)=666.06318;O4(1,6)=666.06318;O4(1,7)=666.06318;
O4(1,8)=666.06318;O4(1,9)=11796.029;O4(1,10)=666.06318;O4(2,1)=700.04653;O4(2,2)=700.04653;O4(2,3)=700.04653;O4(2,4)=11210.11;
O4(2,5)=11210.11;O4(2,6)=11210.11;O4(2,7)=11210.11;O4(2,8)=11210.11;O4(2,9)=686.96155;O4(2,10)=11210.11;O4(3,1)=11787.847;
O4(3,2)=11787.847;O4(3,3)=11787.847;O4(3,4)=653.16005;O4(3,5)=653.16005;O4(3,6)=653.16005;O4(3,7)=653.16005;O4(3,8)=653.16005;
O4(3,9)=45.00978;O4(3,10)=653.16005;

O3(1,1)=0.0036614;O3(1,2)=0.0036614;O3(1,3)=0.0036614;O3(1,4)=0.0526316;O3(1,5)=0.0526316;O3(1,6)=0.0526316;O3(1,7)=0.0526316;
O3(1,8)=0.0526316;O3(1,9)=0.9409864;O3(1,10)=0.0526316;O3(2,1)=0.0553521;O3(2,2)=0.0553521;O3(2,3)=0.0553521;O3(2,4)=0.8947368;
O3(2,5)=0.8947368;O3(2,6)=0.8947368;O3(2,7)=0.8947368;O3(2,8)=0.8947368;O3(2,9)=0.0553521;O3(2,10)=0.8947368;O3(3,1)=0.9409864;
O3(3,2)=0.9409864;O3(3,3)=0.9409864;O3(3,4)=0.0526316;O3(3,5)=0.0526316;O3(3,6)=0.0526316;O3(3,7)=0.0526316;O3(3,8)=0.0526316;
O3(3,9)=0.0036614;O3(3,10)=0.0526316;

O2(1,1)=0.0038911;O2(1,2)=0.0038911;O2(1,3)=0.0038911;O2(1,4)=0.0588235;O2(1,5)=0.0588235;O2(1,6)=0.0588235;O2(1,7)=0.0588235;
O2(1,8)=0.0588235;O2(1,9)=1.;O2(1,10)=0.0588235;O2(2,1)=0.0588235;O2(2,2)=0.0588235;O2(2,3)=0.0588235;O2(2,4)=1.;O2(2,5)=1.;
O2(2,6)=1.;O2(2,7)=1.;O2(2,8)=1.;O2(2,9)=0.0588235;O2(2,10)=1.;O2(3,1)=1.;O2(3,2)=1.;O2(3,3)=1.;O2(3,4)=0.0588235;O2(3,5)=0.0588235;
O2(3,6)=0.0588235;O2(3,7)=0.0588235;O2(3,8)=0.0588235;O2(3,9)=0.0038911;O2(3,10)=0.0588235;

O1(1,1)=0.0038911;O1(1,2)=0.0038911;O1(1,3)=0.0038911;O1(1,4)=0.0588235;O1(1,5)=0.0588235;O1(1,6)=0.0588235;O1(1,7)=0.0588235;
O1(1,8)=0.0588235;O1(1,9)=1.;O1(1,10)=0.0588235;O1(2,1)=0.0588235;O1(2,2)=0.0588235;O1(2,3)=0.0588235;O1(2,4)=1.;O1(2,5)=1.;
O1(2,6)=1.;O1(2,7)=1.;O1(2,8)=1.;O1(2,9)=0.0588235;O1(2,10)=1.;O1(3,1)=1.;O1(3,2)=1.;O1(3,3)=1.;O1(3,4)=0.0588235;O1(3,5)=0.0588235;
O1(3,6)=0.0588235;O1(3,7)=0.0588235;O1(3,8)=0.0588235;O1(3,9)=0.0038911;O1(3,10)=0.0588235;

call BACKWARD(E,Y,p,L,k,pqr,O5,er5,er4,er3)
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After coping this i realize that i haven't defined the er3 variable in the main code. I feel so stupid
now it works fine. Thank anyway
I'm still a beginner so if you know of a better way to write this code i really appreciate some constructive criticism.