n(x)

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Elias Hassan Rezai
Elias Hassan Rezai 2020 年 10 月 25 日
コメント済み: Elias Hassan Rezai 2020 年 10 月 25 日
can someboduy tell me wqhat is does these three kind of code mean? I mean why code3 is not working and what dows a( r ) actually mean? hwo does these change the layout? by rthe way this is not a homework but only my own thought :)
CODE1)
a=[0 1];
for r=3:100
a(r)=(r-1)+(r-2);
end
disp(a)
CODE2)
a=[0 1];
for r=3:100
a(r)=a(r-1)+a(r-2);
end
disp(a)
CODE3)
a=[0 1];
for r=3:100
a(r)=a(r-1)+a(r-2);
end
disp(a)
  2 件のコメント
Walter Roberson
Walter Roberson 2020 年 10 月 25 日
? Code 3 is the same as Code 2 and produces the same result.
Elias Hassan Rezai
Elias Hassan Rezai 2020 年 10 月 25 日
I forgot to change that; I meant a=(1) in code 3 :)

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採用された回答

drummer
drummer 2020 年 10 月 25 日
編集済み: drummer 2020 年 10 月 25 日
CODE1)
a=[0 1]; % you tell MATLAB a is a vector
for r=3:100 % the for loop you understand, right?
a(r)=(r-1)+(r-2); % you're attributing to the rth term of your vector a values of (r - 1) + (r - 2)
% the 1st iteration will be a(3) = (3 - 1) + (3 - 2); a(4) = (4 - 1) + (4 - 2); so on...
end
disp(a)
CODE2)
a=[0 1]; %from this line, a(1) = 0 and a(2) = 1.
for r=3:100
a(r)=a(r-1)+a(r-2); % the difference between this and code 1) is that you're performing
% the summation calling the indexes in the vector a, rather than r values.
% so a(3) = a(2) + a(1) = 0 + 1 = 1
% In code 1, a(3) = 2 + 1 = 3
end
disp(a)
Code 3) is the same as 2) Walter mentioned.
Hope that helps...
Cheers
  1 件のコメント
Elias Hassan Rezai
Elias Hassan Rezai 2020 年 10 月 25 日
thank u so much :)))))

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