# coverting fortran to matlab

1 view (last 30 days)
hasan damaj on 29 Nov 2019
Answered: hasan damaj on 11 Dec 2019
hasan damaj on 29 Nov 2019
ok no problem.
line 22 to line 28.
but i need ur help to continue the code and thank u
hasan damaj on 29 Nov 2019
please continue this code program for me..knowing i will attach the picture for the results and thank you

J Chen on 29 Nov 2019
The following statements are wronng
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;
It should be
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 );
Change while amax > 0.01 to
while 1
..
if ( amax > 0.01 )
amax = 0;
else
break
end
end
One end statement has been missing in your code.
dpb on 29 Nov 2019
Edited: dpb on 29 Nov 2019
Where is the looping structure in the original? That's what you need to emulate.
I don't quite agree with the other poster's suggestion (not that it won't work but it isn't my "cup of tea" in how I'd write it).
I pointed out above the modifications needed to write the while as
amax=1;
while amax>E % set E to desired tolerance value
...
end
Walter Roberson on 29 Nov 2019
while 1 .... ???????
.. (what is this)??????
In MATLAB, if and while are considered true provided that all of the values in the condition are non-zero. The constant 1 there will always be non-zero, so this code is expressing an infinite loop.
MATLAB happens to use the numeric value 1 for the logical value true so a directly equivalent way of writing while 1 is while true -- which is a form I am more likely to write.

hasan damaj on 11 Dec 2019
h=zeros(nx,ny);
amax = 0;
h(1,:) = [ 8.04 , 7.68 , 7.19 , 8.53];
h(end,:) = [6.82 , 7.56 , 7.99 , 8.29 ];
h(2:3,1) = [7.68;7.19];
h(2:3,end) = [8.41; 8.33];
while amax >= 0.01
for j = 2 : nx-1
for i = 2 : ny-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
h
its giving me zeros in h(2,3) h(3,2) h(3,3) h(2,2)
how can i get to the results same as above

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