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Error: File: square_matrix.m Line: 12 Column: 16

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Bennie
Bennie 2012 年 2 月 27 日
Hello I am trying to apply the rule:(S(i,j))^t = (S(i,j-1))^t+ (S(i-1,j))^t+ (S(i,j))^t + (S(i+1,j))^t + (S(i,j+1))^t
I am getting an error for entering the matrix incorrectly. Below is my full code. clear; clc; clf; disp('Starting...');
S=zeros(50);
newS=zeros(50);
%i=0:5;
%j=0:5;
for t=1:200
for i= 1:5
j= 1:5
(S(i,j))^t = (S(i,j-1))^t+ (S(i-1,j))^t+ (S(i,j))^t + (S(i+1,j))^t + (S(i,j+1))^t;
if 0<=S(i,j)< 3
then S(i,j)=0
elseif 3<= S(i,j)<= 5
then S(i,j)=1
end
end
newS=S(i,j)
end
spy(newS)
actual error when run: ??? Error: File: square_matrix.m Line: 12 Column: 16 The expression to the left of the equals sign is not a valid target for an assignment.
Please advise.

回答 (1 件)

Jan
Jan 2012 年 2 月 27 日
What do you expect "^t" to do? It means "power to t". If you want to transpose a matrix, swap the indices instead.
The error message means, that you cannot apply a function on the left hand side of an assignment.
After you have fixed this, the next problems will appear:
if 0<=S(i,j)< 3
In Matlab this calculates:
S(i,j) < 3 ==> TRUE or FALSE
0 <= TRUE or FALSE
which is TRUE always, because TRUE is evaluated as 1 and FALSE as 0. You want:
if 0 <= S(i,j) && S(i,j) < 3
In addtion there is no then keyword in Matlab.
What are you expecting as result of these lines:
for i= 1:5
j= 1:5
?
The line:
newS=S(i,j)
overwrites the matrix newS in the first iteration and replaces the scalar value in all subsequent iterations. Therefore newS is a scalar at the end. Do you want newS(i, j)?
Finally I think the shown code is far away from a valid Matlab syntax. I suggest to read the Getting Started chapters of the documentation to learn the basics of this powerful programming language.

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