kindly help in the conversion of fortran code to matlab
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Oluwaseyi Aliu
am 10 Apr. 2020
Kommentiert: Oluwaseyi Aliu
am 10 Apr. 2020
5 Kommentare
Walter Roberson
am 10 Apr. 2020
snul = 0.0;
snur = 0.0;
for j = 0:m
rnul = (th(0,j) - th(1,j)) * n;
rnur = (th(n-1,j) - th(n,j)) * n;
snul = snul + rnul;
snur = snur + rnur;
fprintf('%g %g %g\n' j/m, rnul, rnur);
end
avnl = snul / m;
avnr = snur / m;
fprintf('%g %g %g\n', ra, avnl, avnr);
I do not see any 100 x 100 matrices, not unless you are saying that th is a matrix. In fortran it is possible but not all that common to declare arrays to start from index 0, or any other index. I would want to see the statements that defined th, such as
REAL*8 th(100,100)
but it could not be exactly like that if th is a matrix being indexed at 0.
Akzeptierte Antwort
Walter Roberson
am 10 Apr. 2020
n = 100; m = 100;
c8 = length(0:8);
f = zeros(c8, n+1, m+1);
feq = zeros(c8, n+1, m+1);
rho = zeros(c8, n+1, m+1);
w = zeros(1, c8);
cx = zeros(1, c8);
cy = zeros(1, c8);
u = zeros(n+1, m+1);
v = zeros(n+1, m+1);
g = zeros(c8, n+1, m+1);
geq = zeros(c8, n+1, m+1);
th = zeros(n+1, m+1);
i = 0;
%then some code to read the data.
%then
snul = 0.0;
snur = 0.0;
for j = 0:m
rnul = (th(1+0,1+j) - th(1+1,1+j)) * n;
rnur = (th(1+n-1,1+j) - th(1+n,1+j)) * n;
snul = snul + rnul;
snur = snur + rnur;
fprintf('%g %g %g\n' j/m, rnul, rnur);
end
avnl = snul / m;
avnr = snur / m;
fprintf('%g %g %g\n', ra, avnl, avnr);
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Fortran with MATLAB finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!