Réponse acceptée

Azzi Abdelmalek
Azzi Abdelmalek le 26 Avr 2014

1 vote

a||b will return 1 if the first expression a is true, without evaluating the second expression b
Example
2==2 || hhh % even hhh is not defined Matlab will not evaluate it, because the first expression 2==2 is true
a&&b will return 0 if the first expression a is false without evaluating the second expression b

6 commentaires

john
john le 26 Avr 2014
OK Azzi,
what about a|b? What is different?
Azzi Abdelmalek
Azzi Abdelmalek le 26 Avr 2014
Modifié(e) : Azzi Abdelmalek le 26 Avr 2014
a | b and a||b should give the same result, the difference is: Matlab when evaluating a|b, a and b are both evaluated, try
2==2 | h
In this case the first expression is 2==2, which is true, we know that if one of the two expression is true, then the result is true, without evaluating the second expression, but by using | Matlab will evaluate the second expression which is h, in our case h is not defined, that's why you will get an error.
john
john le 26 Avr 2014
ok, thank you
john
john le 26 Avr 2014
but result :
if 2==2 || h
fff=1
end
is the same like:
if 2==2 | h
fff=1
end
but for this I get error:
if h | 2==2
fff=1
end
Azzi Abdelmalek
Azzi Abdelmalek le 26 Avr 2014
Modifié(e) : Azzi Abdelmalek le 26 Avr 2014
Because Matlab, in both case (| or | | ) evaluate the first expression h which is not defined.
Azzi Abdelmalek
Azzi Abdelmalek le 26 Avr 2014
It's better to use | |, this can make your code faster, when the first expression is true, Matlab doesn't need to evaluate the second one.

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Plus de réponses (1)

dpb
dpb le 26 Avr 2014

1 vote

Depends entirely on the purpose...the double logical operators short-circuit and return only a scalar whereas the single ones are point-by-point operators over the full dimension of the two operands and return a matrix of the same size.
doc relop
has further details and info

2 commentaires

john
john le 26 Avr 2014
a=3;
b=2;
What is better ?
if a==3 || b==2
end;
Or
if a==3 | b==2
end;
Jan
Jan le 27 Avr 2014
Modifié(e) : Jan le 27 Avr 2014
@John: When a and b are scalars, both versions are equivalent. But the first one || is slightly faster (nano-seconds for scalar operands...), when the first expression is true already. When a and/or b is a vector, you need the , which is equivalent to |or(a==3, b==2). But then the vector expression in the if command is tricky, because implicitly this is performed:
expr = or(a==3, b==2);
if all(expr) && ~isempty(expr) ...
This is at least confusing or can even be a bug, if this behavior is not intended.

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