double conditional in one line

By chance I tried a sentence like

0<x<1

i cannot find any documentation for this.

At the beginnnig looks like weird

>> x=0.23: 0<x<1

ans =

logical

0

but

>> x=0.23: 0.1<x<1.1

ans =

logical

1

You can concatenate more, exemple

>> x=1; y=2; 0<x<y<2

ans =

logical

1

I still cannot figure out how it really works. Any help?

Answer 1

[Steven Lord]

At the beginnnig looks like weird

>> x=0.23: 0<x<1

This does not ask the question "Is x between 0 and 1 exclusive?" It is instead interpreted as:

x = 0.23;
y = (0 < x) < 1

y = logical
   0

The first part of the expression for y, (0 < x), returns either false (treated as 0) or true (treated as 1) depending on the value you specified for x. In this case since 0 is less than 0.23, it is true. So y becomes:

y = true < 1

y = logical
   0

Since true is treated as 1 and 1 is not less than 1, y is false.

Your second expression, x=0.23: 0.1<x<1.1, will always return true. Both false (0) and true (1) are strictly less than 1.1 so it doesn't matter whether 0.1 is less than x or not.

x = 0.23;
z = 0.1<x<1.1

z = logical
   1

z = (0.1 < x) < 1.1

z = logical
   1

z = true < 1.1

z = logical
   1

x = Inf;
z = 0.1<x<1.1

z = logical
   1

If you enter this in the Editor, Code Analyzer should warn about that pattern and suggest the approach others have already recommended earlier, using two separate tests combined using the & or && operators.

x = 2;
z = (0.1 < x) & (x < 1.1)

z = logical
   0

[Antonio Baeza]

Thank you!. it is clear now

Answer 2

[Matt J]

You need to be using '&'

 x=0.23; 
 0<x & x<1

ans = logical
   1

Answer 3

[Les Beckham]

You have to join the multiple conditions with a logical operator. For example, the condition 0<x<1 is written like this

x=0.23;
0<x && x<1

ans = logical
   1

and this one 0<x<y<2 is written like this

x=1;
y=2;
0<x && x<y && y<2

ans = logical
   0

[Antonio Baeza]

Oh, yes, I know how to make a regular conditional. I was junt wondering how the double one works. Many thanks @Les Beckham and @Matt J

[Les Beckham]

You are quite welcome.