what's wrong with the code

for i=1:n
  for j=1:n
      if (5<i<15 & 8<j<12 & j=i) {d(i,j)=1};
      else d(i,j)=0;
      end
  end
end

[Youssef Khmou]

Identity matrix In?

Answer 1

[Matt J]

Of course, the whole thing could be done much more simply and without loops,

 z=zeros(1,n);
 z(min(9:11,n))=1;

 d=diag(z);

Answer 2

[Azzi Abdelmalek]

n=40
for i=1:n
  for j=1:n
      if (5<i & i<15 & 8<j & j<12 & j==i)  d(i,j)=1;
      else d(i,j)=0;
      end
  end
end

Answer 3

[Youssef Khmou]

modify,

for i=1:n
for j=1:n
if (5<i<15 && 8<j<12 && j==i)
d(i,j)=1;
else d(i,j)=0;
end
end
end

[Walter Roberson]

This has the same bug as the original. 5<i<15 means ((5<i)<15) which means "(true (1) or false (0)) < 15" which is always true.