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how to convert to matlab from fortran

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hasan damaj
hasan damaj on 25 Dec 2019
Commented: hasan damaj on 26 Dec 2019

  1 Comment

hasan damaj
hasan damaj on 25 Dec 2019
continuation of the code is attachedco2.JPG

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Answers (2)

the cyclist
the cyclist on 25 Dec 2019
You could try f2matlab. It might not get you all the way to the final code you need, but it will probably help.

  6 Comments

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hasan damaj
hasan damaj on 25 Dec 2019
Thank you for your comment.
I'm trying to convert the Fortran code to MATLAB.
"""""this is a regional flow example using gauss-seidel iteration"""
Image Analyst
Image Analyst on 25 Dec 2019
That is supposed to be a comment. In MATLAB, it would look like this:
% This is a regional flow example using Gauss-Seidel iteration.
hasan damaj
hasan damaj on 25 Dec 2019
i know im not writing a code here :)
thank you

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Ben Barrowes
Ben Barrowes on 26 Dec 2019
OCR and some hand editing produced this fortran code which I compiled and ran:
program flow
C RE GIONAL FLOW SYST&M &XAMPLE
dimENSION H(13, 7)
C INITIALIZE ALL H(I,.JJ VALUES TO BE l00.
DO 5 J=1,7
do 5 I=1,l3
H(I,J) = 100.
5 CONTINUE
c WATER TABLE BOUNOARY
DX=28.
DO 10 i=2,12
H(I,1) = 0.02*DX*(I-2)-100.
10 CONTINUE
C KEEP TRACK OP NUMBER OF ITERATIONS AND OF larRGEST ERROR
C NO-FLOW doundaries NEeD TO BE RESET WITHIN EACH ITERATION LOOP
NUMIT = 0
35 AMAX =0.
NUmIT = NUMIT + 1
C LeFT AND RIGHT NO-PLOW BOUNDARIES
DO 20 J=1,7
H(1,J) = H(3,J)
H (13,J) = H(11,J)
20 COnTINUe
C BOT10M N!rtLCW SOUNDAFC:t
DO 30 i=2,12
H(i,7) =H(i,5)
30 CONTINUE
C SWEEP INTERIOR POINTS WITH 5-POltn' OPERATOR
DO 40 J=2,6
DO 40 i=2,12
OLDVAL= H(I,J)
H(i,J) = (H(I-1,J) + H(I+1,J) + H(I,J-1) + H(I,J+1))/4.
ERR= ABS(H(I,J) -OLdVAL)
IF(ERR.GT.AKAX) AMAX=ERR
40 CONTINUE
C 00 ANOTHER ITERATION IF LARGEST ERROR AFFECTS JRD DECIMAL PLACE
IF (AMAX.GT.0.001) GO TO 35
C WE ARE OONE.
PRINT 50,NUmIT, ((H(I,J),I=2,12),J=l,6)
50 formAT(///1X,'NUMBER OF ITERATIONS IS',I4,///6(11f8.2///))
ENd
Using f2matlab and some other tools I have, this is the resulting matlab code. It seems to run, but I have not tested for accuracy of results. The fprintf statement could use some fixing for the results to look exactly the same.
function flow(varargin)
clear global;
clear functions;
global GlobInArgs nargs
GlobInArgs={mfilename,varargin{:}};
nargs=nargin+1;
global unit2fid;
if ~isempty(unit2fid), unit2fid={};
end
persistent akax amax dx err firstCall h i j l l3 numit oldval format_50
;
if isempty(firstCall),firstCall=1;end;
if firstCall;
format_50=[ '\n' , '\n' , '\n' ,blanks(1),'NUMBER OF ITERATIONS IS','%4d', '\n' , '\n' , '\n' ,'~',repmat([repmat('%8.2f',1,11), '\n' , '\n' , '\n' ] ,1,6)];
akax=0;
amax=0;
dx=0;
err=0;
h=zeros(13,7);
i=0;
j=0;
l=0;
l3=0;
numit=0;
oldval=0;
end
firstCall=0;
% RE GIONAL FLOW SYST&M &XAMPLE
% INITIALIZE ALL H(I,.JJ VALUES TO BE l00.
for j = 1: 7
for i = 1: l3
h(i,j) = 100.;
end
i = fix(l3+1);
end
j = fix(7+1);
% WATER TABLE BOUNOARY
dx = 28.;
for i = 2: 12
h(i,1) = 0.02.*dx.*(i-2) - 100.;
end
i = fix(12+1);
% KEEP TRACK OP NUMBER OF ITERATIONS AND OF larRGEST ERROR
% NO-FLOW doundaries NEeD TO BE RESET WITHIN EACH ITERATION LOOP
numit = 0;
while (1);
amax = 0.;
numit = fix(numit + 1);
% LeFT AND RIGHT NO-PLOW BOUNDARIES
for j = 1: 7
h(1,j) = h(3,j);
h(13,j) = h(11,j);
end
j = fix(7+1);
% BOT10M N!rtLCW SOUNDAFC:t
for i = 2: 12
h(i,7) = h(i,5);
end
i = fix(12+1);
% SWEEP INTERIOR POINTS WITH 5-POltn' OPERATOR
for j = 2: 6
for i = 2: 12
oldval = h(i,j);
h(i,j) =(h(i-1,j)+h(i+1,j)+h(i,j-1)+h(i,j+1))./4.;
err = abs(h(i,j)-oldval);
if(err > akax)
amax = err;
end
end
i = fix(12+1);
end
j = fix(6+1);
% 00 ANOTHER ITERATION IF LARGEST ERROR AFFECTS JRD DECIMAL PLACE
if(amax > 0.001)
continue;
end
% WE ARE OONE.
h(2:12,1:6)
%fprintf(1,[format_50],numit,{{h(i,j),'i','2','1','12'},'j','l','1','6'});
tempBreak=1;
break;
end
clear all
end %program flow

  5 Comments

Show 2 older comments
hasan damaj
hasan damaj on 26 Dec 2019
Mr. Ben, i attached a possible code i figured can u just look at it,
clc
clear
nx = 7;
ny = 13;
h = zeros(nx,ny);
h(:,:) = 100;
h;
Dx = 20;
amax = 1;
% while amax >= 0.001
% amax = 0;
for j = 2 : ny-1
h(1,j) = 0.02 * Dx * (j-2) + 100;
for i = 1 : nx
h(i,1) = h(i,3);
h(i,13) = h(i,11);
for j = 2 : ny-1
h(7,j) = h(5,j);
while amax >= 0.001
amax = 0 ;
for i = 2 : ny - 1
for j = 2 : nx - 1
oldval = h(i,j);
h(i,j) = (h(i-1,j) + h(i+1,j) + h(i,j-1) + h(i,j+1))/4;
e = abs(h(i,j) - oldval);
if e > amax
amax = e ;
end
end
end
end
end
end
end
h
is the location of while loop ok?? its giving me index exceeds matrix dimensions.
the above comments shows you the result of table of numbers
thank you
Ben Barrowes
Ben Barrowes on 26 Dec 2019
One thing I noticed looking at your code is that you have reused the same variable as the loop index variables. You have three loops that use J as the loop variable for example and two that use I as the loop variable. Use different variables for each loop index.
hasan damaj
hasan damaj on 26 Dec 2019
i cant change the variable because its 1 equation where h is in terms of i and j

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