필터 지우기
필터 지우기

Call variables in for loop

조회 수: 7 (최근 30일)
Rok mr
Rok mr 2012년 3월 22일
Hi to all,
L1=4 L2=3 ... L30=12
for i=1:30 P(:,i)=sprintf(L%1d,i) - sprintf(L%1d,i+1) end
The result should be P(1)=L1-L2, P(2)=L2-L3, P(3)=L3-L4, P(4)=L4-L5 but it's not :) What should I use instead sprintf?
Thank you for you answers!!!

채택된 답변

Kevin Holst
Kevin Holst 2012년 3월 22일
To do this like you're attempting requires the use of the eval function, but be very careful in its use. Google 'eval matlab' for more information on why. And actually, the way you've got it set up wouldn't work. Your for loop needs to run from 1:29 because there's no L31.
for i = 1:29
eval(['P(' num2str(i) ')=L' num2str(i) '-L' num2str(i+1) ';'])
end
Is there any reason why your L variables aren't in a single L vector?
  댓글 수: 7
Kevin Holst
Kevin Holst 2012년 3월 22일
@Oleg, I haven't tried the eval(eval(eval(... suggestion, but I did evaluate char([102 108 105 112 108 114 40 39 116 105 120 101 39 41]) in my workspace... very sneaky! ;) That's the nice way to show a problem with eval. The mean way would be to "accidentally" have someone recursively remove all of the contents of their hard drive. That'd be a bad day.
Oleg Komarov
Oleg Komarov 2012년 3월 22일
Should have concealed it better...

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Whos에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by