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

1 Ansicht (letzte 30 Tage)
yang-En Hsiao
yang-En Hsiao am 31 Mär. 2019
Bearbeitet: madhan ravi am 31 Mär. 2019
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

Antworten (1)

madhan ravi
madhan ravi am 31 Mär. 2019
Bearbeitet: madhan ravi am 31 Mär. 2019
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?
  5 Kommentare
yang-En Hsiao
yang-En Hsiao am 31 Mär. 2019
ok i understand,but why will the matlab stop and warn when when the x3 is 27?
madhan ravi
madhan ravi am 31 Mär. 2019
Bearbeitet: madhan ravi am 31 Mär. 2019
% 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?

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by