Effacer les filtres
Effacer les filtres

Why is the if statement not working?

2 vues (au cours des 30 derniers jours)
tash7827
tash7827 le 3 Août 2015
this is my code - it is supposed to be a binary array, where the length of time (through the array) that the "1"s are held is determined by "time" - why is the if statement not working based off of the output of "time"?
function cube=cube(exp)
exp=zeros(60,96,10);
time=randi([3 7]);
if time==3
for j=1:size(exp,1);
for jj=1:size(exp,2);
rand=randi([1 8]);
exp(j,jj,rand:rand+2)=1;
end
end
elseif time == 4
for j=1:size(exp,1);
for jj=1:size(exp,2);
rand=randi([1 7]);
exp(j,jj,rand:rand+3)=1;
end
end
elseif time == 5
for j=1:s ize(exp,1);
for jj=1:size(exp,2);
rand=randi([1 6]);
exp(j,jj,rand:rand+4)=1;
end
end
elseif time == 6
for j=1:size(exp,1);
for jj=1:size(exp,2);
rand=randi([1 5]);
exp(j,jj,rand:rand+5)=1;
end
end
elseif time == 7
for j=1:size(exp,1);
for jj=1:size(exp,2);
rand=randi([1 4]);
exp(j,jj,rand:rand+6)=1;
end
end
cube=exp
end
  1 commentaire
the cyclist
the cyclist le 3 Août 2015
One suggestion, aside from trying to solve your main problem, is that you should not be using MATLAB keywords as variable names, because the code may behave in unexpected ways, and also the code simply becomes very confusing (especially for veteran users). In your case, exp and rand should be avoided as variable names.

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Réponse acceptée

the cyclist
the cyclist le 3 Août 2015
One problem is that in the elseif statement where you check if time == 5, you have
for j=1:s ize(exp,1);
instead of
for j=1:size(exp,1);

Plus de réponses (1)

the cyclist
the cyclist le 3 Août 2015
Modifié(e) : the cyclist le 3 Août 2015
You need
time==3
instead of
time=3
== is the "determine equality" operator. = is the assignment statement.
  1 commentaire
tash7827
tash7827 le 3 Août 2015
sorry -- i hadn't realized this was old code - i already have what you suggested and it still is not working - my whole array is still all zero.

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