# What is the output?

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Shivangi Srivastava 2018 年 4 月 1 日
コメント済み: Walter Roberson 2018 年 4 月 1 日
I am trying to check how the loop works hence, I am displaying the value of n in the for loop but it doesn't seem to work. Why doesn't n get displayed and how is the code (loop) working?
h=0.3;
x=0:h:0.8;
for n=4:length((x)-1)
disp(n)
% Doing something here
end

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### 採用された回答

Walter Roberson 2018 年 4 月 1 日
length of x is 3. When you subtract 1 from a numeric vector you do not change the length so length((x) - 1) is the same length(x) which is 3. So your for loop is
for n=4:3
Which will not execute at all.

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### その他の回答 (1 件)

Roger Stafford 2018 年 4 月 1 日
The vector x has only three elements, [0,.3,.6], so the 'for' instruction reads:
for n = 4:2
Therefore the loop will not execute at all, and this accounts for your lack of results.
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Shivangi Srivastava 2018 年 4 月 1 日

If h would have been 0.2 then the for loop would be n = 4:4 right? How does this code get executed and the values of n change?
Walter Roberson 2018 年 4 月 1 日
No, with 0.2 x would be 0 0.2 0.4 0.6 0.8 which is length 5. When you subtract one from the numeric values the length stays the same, just you would working with -1 -.8 -.6 -.4 -.2. Still length 5. So the loop would be 4:5

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