why does the window show me x3=27 when x3=1:2:50?

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yang-En Hsiao
yang-En Hsiao 2019-3-31
编辑： madhan ravi 2019-3-31
I wrote this code ,and the c3 can be a vector ,the elements in this vector is the result of every loops,however,i found that the window show me an error ,here is my code
XX3=1:2:50
AXX3=zeros(1,length(XX3))
a3=5
b3=-3
for x3=1:2:50
y3=a3*x3^2+b3
c3(x3)=AXX3(x3)+y3
end
And here is the error
Attempted to access AXX3(27); index out of bounds
because numel(AXX3)=25.
Error in wqeqwe (line 7)
c3(x3)=AXX3(x3)+y3
I found that the x3 = 27.i wonder why? x3 is 1:2:50.there should be only 25 elements,i mean ,x3 should be as same as XX3,does anyone know where is my mistake

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回答（1 个）

madhan ravi
madhan ravi 2019-3-31
编辑：madhan ravi 2019-3-31
No loop's needed:
XX3=1:2:50;
a3=5;
b3=-3;
x3=1:2:50;
y3=a3*x3.^2+b3;
AXX3=zeros(size(XX3));
c3=AXX3+y3; % if AXX3 is zero then why add it with y3 to get c3?
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yang-En Hsiao
yang-En Hsiao 2019-3-31
ok i understand,but why will the matlab stop and warn when when the x3 is 27?
madhan ravi
madhan ravi 2019-3-31
编辑：madhan ravi 2019-3-31
Jeez..,What is Indexing and why it is the first and foremost important thing to learn before making tasks difficult?
% example
x = 1:25 % 25 elements
x(27) % will throw error why?
% ^^---- there exists no such element in the 27th place in x because x has only 25 elements not 27, get it?

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