Why is the string below the first conditional not printed?

I need help understanding why the string below the first conditional is not printed when AA, BB, CC are clearly equal. See attachment below

2 件のコメント

dpb
dpb 2021 年 11 月 13 日
(AA==BB) == 1 && (CC) == 1
is not good syntax.
if AA==1 & BB==1 & CC==1
will work; more succinct forms can be written if you would use arrays for variables instead of indivdually naming the variables (generally a bad practice to use MATLAB efficiently to have mulitple similarly-named/functionally equivalent variables as above).
Andromeda
Andromeda 2021 年 11 月 14 日
I tried if (AA == BB) == (CC ==1) first and that didn't produce the expected output and thus resorted to what you see above. Why doesn't if (AA == BB) == (CC ==1) work as well?
"if you would use arrays for variables instead of indivdually naming the variables ", care to give me an example?

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回答 (1 件)

Cris LaPierre
Cris LaPierre 2021 年 11 月 14 日
編集済み: Cris LaPierre 2021 年 11 月 14 日

1 投票

Because the result of the dot products is not exactly 1.
A = [2 2 -1+4i]/5;
B = circshift(A,-1);
C = circshift(B,-1);
AA = dot(A,A)
AA = 1.0000
BB = dot(B,B)
BB = 1.0000
CC = dot(C,C)
CC = 1.0000
% now subtract 1. Result is not 0
AA-1
ans = 2.2204e-16
BB-1
ans = 2.2204e-16
CC-1
ans = 2.2204e-16

2 件のコメント

Andromeda
Andromeda 2021 年 12 月 2 日
Well, that is odd. For a vector to form an orthonormal basis[algebra], the dot product of the vector with itself should be equal to 1
dpb
dpb 2021 年 12 月 2 日
Floating point arithmetic is only an approximation to algebra...see <Goldberg - What Every Computer Scientist Should Know> for detailed explanation. There are some other discussions and examples in the documentation as well.

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