Why does my resulting matrix only contain the last value of the iteration for p and q?

phi(1,1) = 3;

phi(2,1) = 2;

phi(3,1) = 1;

phi(5,1) = 6;

g = [1 2 3; 4 5 6; 7 8 9];

for j = [1,2,5]

      for m = [1,2,5]

          for p = 1:3

              for q = 1:3

                  k(phi(j),phi(m)) = g(p,q);

              end

          end

      end
  end

Answer 1

[Walter Roberson]

Your destination

k(phi(j),phi(m))

does not change as p or q change, so each iteration of p and q, you are writing to the same output location, so the result is the same as if you had only done the final assignment.

[Celso Carranza]

If I were to relate j and m with p and q, for instance

phi(1,1) = 3;

phi(2,1) = 2;

phi(3,1) = 1;

phi(5,1) = 6;

g = [1 2 3; 4 5 6; 7 8 9];

for j = [1,2,5]

      for m = [1,2,3]

          p = j;

          q = m;

                  k(phi(j),phi(p)) = g(m,q);
      end
  end

The result still ends up if I were only doing the final assessment.

[Walter Roberson]

You have

p = j

so when you do

k(phi(j),phi(p))

then that is the same as

k(phi(j), phi(j))

and as j is your outer loop, you are still ovewriting the same location for each m value.