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this should be easy I have a while loop that is infinite

3 ビュー (過去 30 日間)
SonOfAFather
SonOfAFather 2012 年 9 月 13 日
When I run this it is inifinite and well that's not what i am looking for... i thought that by limiting the x value to 99 which is in the vector i could get it to stop but that's not the case. any ideas i know that it is going to be something simple...
clc, clear all, close all;
x = 1;
while x <= 99
x = 1:2:100
end
  2 件のコメント
SonOfAFather
SonOfAFather 2012 年 9 月 13 日
I see that i can use the break comand, but don't think that is the best choise here anyother options out there??
SonOfAFather
SonOfAFather 2012 年 9 月 13 日
what do you guys think of this changing the while x<=99 to x==1? i found that the loop stops as i would want it to.

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採用された回答

Matt Fig
Matt Fig 2012 年 9 月 13 日
In MATLAB, conditionals evaluate to true if ALL values of the conditional are non-zero or true. You are repeatedly creating a vector x inside the loop which has all of its values equal to or less than 99, so the conditional will evaluate to true every time. This is why you are getting an infinite loop.
Here is an example, using an IF statement. It evaluates the same way the WHILE loop does:
v = [1 2 -9]; % Notice ALL values are non-zero
if v, disp('IN IF 1'),end
v = [1 0 -9]; % Not all values are non-zero.
if v, disp('IN IF 2'),end
When you do a comparison (>,<,==, etc), you get a logical result. If you do a comparison on an array, you get a logical vector the same size as the array. This will go through a conditional expression the same as v did above: any false (0) values and the expression will not pass the statement.
v>-20 % This will pass [1 1 1]
v<0 % Will not pass, [0 0 1]
What are you trying to achieve with your loop?
  5 件のコメント
Matt Fig
Matt Fig 2012 年 9 月 13 日
編集済み: Matt Fig 2012 年 9 月 13 日
I see. There are several ways to do that.
x = [];
n = 1;
while n<=100
x = [x n];
n = n + 2;
end
Another:
x = zeros(1,ceil(99/2)); % Not strictly necessary.
n = 1;
while n<=ceil(99/2)
x(n) = 2*n-1;
n = n + 1;
end
You could also be a smart alec:
T = true;
while T
x = 1:2:100;
T = false;
end
SonOfAFather
SonOfAFather 2012 年 9 月 13 日
thanks matt i needed a laugh! :)

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その他の回答 (4 件)

Azzi Abdelmalek
Azzi Abdelmalek 2012 年 9 月 13 日
編集済み: Azzi Abdelmalek 2012 年 9 月 13 日
x = 1;
while x <= 99
x=x+2
% do
end
  4 件のコメント
SonOfAFather
SonOfAFather 2012 年 9 月 13 日
exactly it is a proble that we were given on a study guide i would love to just make it x = 1:2:100 , but the prof wants it to be a while loop.here is the question:
Complete the following problems:
a. Create a for loop that will output the numbers from 1 to 100 by 2s (1, 3, 5, …, 97, 99).
b. Rewrite the above for loop as a while loop.
Azzi Abdelmalek
Azzi Abdelmalek 2012 年 9 月 13 日
x=1
while x<=100-2
x(end+1)=x(end)+2
end

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Wayne King
Wayne King 2012 年 9 月 13 日
編集済み: Wayne King 2012 年 9 月 13 日
Why do you need a while loop here? If you just want a vector running from 1 to 99 in increments of 2, the following does the job.
x = 1:2:100;
With the previous statement you are creating a vector whose last element is 99 and since you don't increment that vector inside the while loop, you never exceed your terminating condition.
note the difference between that and
x = 1;
while x<= 99
x = x+1;
end
x
You see with the above, that when you exit the while loop, x is equal to 100, which terminates the loop. In your example, you never exceed 99 and that is why you have an infinite loop.
  1 件のコメント
SonOfAFather
SonOfAFather 2012 年 9 月 13 日
I need the while loop to because it is what the prof asked for. i agree that the
x = 1:2:100;
is much simpler and cleaner but that's what he wants.
But with 99 being greater-than-or-equal-to x shouldn't 99 be withing the vector since i can see it on the output?

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Wayne King
Wayne King 2012 年 9 月 13 日
If want you create a vector with a while loop (again not sure why you want to do that), there are many ways to code it.
clear x;
n = 1;
k = 1;
while k<100
x(n) = k;
n = n+1;
k = k+2;
end
  3 件のコメント
Azzi Abdelmalek
Azzi Abdelmalek 2012 年 9 月 13 日
Is it a joke?
SonOfAFather
SonOfAFather 2012 年 9 月 13 日
no its an actual problem. here is what i came up with a min ago:
clc, clear all, close all;
x = 0;
for x = 1:2:100;
disp(x)
end
display('while')
x1 = -1;
while x1 <=97;
x1=x1+2;
disp(x1)
end

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Andrei Bobrov
Andrei Bobrov 2012 年 9 月 13 日
x = 1;
while x(end) < 99
x=[x x(end)+2];
end

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