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Koch Square Problem : A(I): index out of bounds; value 4 out of bound 3

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Mohanad Arafae
Mohanad Arafae 2016 年 11 月 14 日
コメント済み: KSSV 2016 年 11 月 15 日
Can someone please fix my error?

回答 (1 件)

Walter Roberson
Walter Roberson 2016 年 11 月 14 日
You use newpointsx(4) and newpointsy(4) before you have assigned anything to those locations. Two lines further on you use the (5) indices too.
  5 件のコメント
Walter Roberson
Walter Roberson 2016 年 11 月 14 日
Your code has
v=[newpointsx(4);newpointsy(4)]-[newpointsx(2);newpointsy(2)]
vprime = R*v
f=[newpointsx(5);newpointsy(5)]-[newpointsx(3);newpointsy(3)]
fprime = R*f
newpointsx(4) = vprime(1)+newpointsx(2) %Point D
newpointsy(4) = vprime(2)+newpointsy(2)
newpointsx(5) = fprime(1)+newpointsx(3) %Point F
newpointsy(5) = fprime(2)+newpointsy(3)
Notice that you use newpointsx(4) and newpointsy(4) in calculating v, but you do not assign any value to newpointsx(4) or newpointsy(4) until 5 lines later in the code.
You need to assign values to newpointsx(4), newpointsx(5), newpointsy(4), newpointsy(5) before you can use them in calculations.
KSSV
KSSV 2016 年 11 月 15 日
@ Walter Roberson, I apologize, I have not seen all this discussion and in hurry closed the question. I shall be careful next time.

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