coverting fortran to matlab

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hasan damaj 2019-11-29

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https://ww2.mathworks.cn/matlabcentral/answers/493855-coverting-fortran-to-matlab

回答： hasan damaj 2019-12-11

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hasan damaj 2019-11-29

ok no problem.

line 22 to line 28.

but i need ur help to continue the code and thank u

hasan damaj 2019-11-29

please continue this code program for me..knowing i will attach the picture for the results and thank you

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回答（2 个）

J Chen 2019-11-29

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https://ww2.mathworks.cn/matlabcentral/answers/493855-coverting-fortran-to-matlab#answer_403951

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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.

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dpb 2019-11-29

编辑：dpb 2019-11-29

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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 2019-11-29

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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.

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hasan damaj 2019-12-11

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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