Interval comparison doesn't bomb in JS
Why doesn't interval comparison bomb in JavaScript?
if(-1 < x < 1)
console.log('x: ', x)
Why are we allowed to do this without getting errors?
Also it seems that:
-1 < x < 1is true forx<=-11 < x < 1is true forx<=1-1 < x < -1is always false-2 < x < 2is always true
In the last 2 cases it seems it is just comparing the 2 ends of the expressions. How are those expressions evalued?
javascript boolean intervals
asked Nov 15 '18 at 13:12
Andrea BergonzoAndrea Bergonzo
5171618
add a comment |
Why doesn't interval comparison bomb in JavaScript?
if(-1 < x < 1)
console.log('x: ', x)
Why are we allowed to do this without getting errors?
Also it seems that:
-1 < x < 1is true forx<=-11 < x < 1is true forx<=1-1 < x < -1is always false-2 < x < 2is always true
In the last 2 cases it seems it is just comparing the 2 ends of the expressions. How are those expressions evalued?
javascript boolean intervals
asked Nov 15 '18 at 13:12
Andrea BergonzoAndrea Bergonzo
5171618
1
They are evaluated as(-1 < x) < 1, which is equivalent to(-1 < x) != trueor!(-1 < x)
– Bergi
Nov 15 '18 at 13:15
What you really need is-1 < x && x < 1
– yunzen
Nov 15 '18 at 13:24
@HerrSerker I know how to do proper comparisons, I was just asking how JS evaluated what in other languages is called interval comparisons.
– Andrea Bergonzo
Nov 15 '18 at 13:59
I knew this. But it would be interesting to other people who want to know, how it's done right
– yunzen
Nov 15 '18 at 14:34
add a comment |
Why doesn't interval comparison bomb in JavaScript?
if(-1 < x < 1)
console.log('x: ', x)
Why are we allowed to do this without getting errors?
Also it seems that:
-1 < x < 1is true forx<=-11 < x < 1is true forx<=1-1 < x < -1is always false-2 < x < 2is always true
In the last 2 cases it seems it is just comparing the 2 ends of the expressions. How are those expressions evalued?
javascript boolean intervals
asked Nov 15 '18 at 13:12
Andrea BergonzoAndrea Bergonzo
5171618
Why doesn't interval comparison bomb in JavaScript?
if(-1 < x < 1)
console.log('x: ', x)
Why are we allowed to do this without getting errors?
Also it seems that:
-1 < x < 1is true forx<=-11 < x < 1is true forx<=1-1 < x < -1is always false-2 < x < 2is always true
In the last 2 cases it seems it is just comparing the 2 ends of the expressions. How are those expressions evalued?
javascript boolean intervals
javascript boolean intervals
asked Nov 15 '18 at 13:12
Andrea BergonzoAndrea Bergonzo
5171618
asked Nov 15 '18 at 13:12
Andrea BergonzoAndrea Bergonzo
5171618
asked Nov 15 '18 at 13:12
Andrea BergonzoAndrea Bergonzo
5171618
asked Nov 15 '18 at 13:12
Andrea BergonzoAndrea Bergonzo
5171618
asked Nov 15 '18 at 13:12
Andrea BergonzoAndrea Bergonzo
5171618
5171618
1
They are evaluated as(-1 < x) < 1, which is equivalent to(-1 < x) != trueor!(-1 < x)
– Bergi
Nov 15 '18 at 13:15
What you really need is-1 < x && x < 1
– yunzen
Nov 15 '18 at 13:24
@HerrSerker I know how to do proper comparisons, I was just asking how JS evaluated what in other languages is called interval comparisons.
– Andrea Bergonzo
Nov 15 '18 at 13:59
I knew this. But it would be interesting to other people who want to know, how it's done right
– yunzen
Nov 15 '18 at 14:34
add a comment |
1
They are evaluated as(-1 < x) < 1, which is equivalent to(-1 < x) != trueor!(-1 < x)
– Bergi
Nov 15 '18 at 13:15
What you really need is-1 < x && x < 1
– yunzen
Nov 15 '18 at 13:24
@HerrSerker I know how to do proper comparisons, I was just asking how JS evaluated what in other languages is called interval comparisons.
– Andrea Bergonzo
Nov 15 '18 at 13:59
I knew this. But it would be interesting to other people who want to know, how it's done right
– yunzen
Nov 15 '18 at 14:34
1
1
They are evaluated as
(-1 < x) < 1, which is equivalent to (-1 < x) != true or !(-1 < x)– Bergi
Nov 15 '18 at 13:15
They are evaluated as
(-1 < x) < 1, which is equivalent to (-1 < x) != true or !(-1 < x)– Bergi
Nov 15 '18 at 13:15
What you really need is
-1 < x && x < 1– yunzen
Nov 15 '18 at 13:24
What you really need is
-1 < x && x < 1– yunzen
Nov 15 '18 at 13:24
@HerrSerker I know how to do proper comparisons, I was just asking how JS evaluated what in other languages is called interval comparisons.
– Andrea Bergonzo
Nov 15 '18 at 13:59
@HerrSerker I know how to do proper comparisons, I was just asking how JS evaluated what in other languages is called interval comparisons.
– Andrea Bergonzo
Nov 15 '18 at 13:59
I knew this. But it would be interesting to other people who want to know, how it's done right
– yunzen
Nov 15 '18 at 14:34
I knew this. But it would be interesting to other people who want to know, how it's done right
– yunzen
Nov 15 '18 at 14:34
add a comment |
3 Answers
3
active
oldest
votes
Because JavaScript allows implicit type coercion, in this case from boolean to number. The -1 < x results in a boolean, which is then implicitly coerced to a number (true = 1, false = 0) for the (result) < 1 part. So:
When
-1 < xis false, the second part is0 < 1which is true.When
-1 < xis true, the second part is1 < 1which is false.
This is covered in the abstract relational comparison algorithm in the spec, and the various operations it links to.
-1 < x < -1 is always false
-2 < x < 2 is always true
In the last 2 cases it seems it is just comparing the 2 ends of the expressions. How are those expressions evalued?
Using x = -1 and x = 1:
- If
x = -1, then-1 < xis false, so the rest is0 < -1, which is false. - If
x = 1, then-1 < 1is true, so the rest is1 < -1which is false. - If
x = -1, then-2 < -1is true, so the rest is1 < -2, which is false. - If
x = 1, then-2 < 1is true, so the rest is1 < -2which is false.
answered Nov 15 '18 at 13:15
T.J. CrowderT.J. Crowder
694k12212381330
add a comment |
I think that this happens because javascript implicitly considers true to have the value 1 and false to have the value 0.
When you do -1 < x < 1, what you're actually doing is (-1 < x) < 1, or true < 1 if x = 0 which is false.
However, if x=-2, (-1 < x < -1) will return true as false < 1 is true. Hope this helps.
You can read more about this here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
answered Nov 15 '18 at 13:20
theapologisttheapologist
576215
add a comment |
The reason is that JS interprets your expression.
if((-1 < x) < 1)
console.log('x: ', x)
Trying using braces...
answered Nov 15 '18 at 13:20
Ram VenkatRam Venkat
787
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Because JavaScript allows implicit type coercion, in this case from boolean to number. The -1 < x results in a boolean, which is then implicitly coerced to a number (true = 1, false = 0) for the (result) < 1 part. So:
When
-1 < xis false, the second part is0 < 1which is true.When
-1 < xis true, the second part is1 < 1which is false.
This is covered in the abstract relational comparison algorithm in the spec, and the various operations it links to.
-1 < x < -1 is always false
-2 < x < 2 is always true
In the last 2 cases it seems it is just comparing the 2 ends of the expressions. How are those expressions evalued?
Using x = -1 and x = 1:
- If
x = -1, then-1 < xis false, so the rest is0 < -1, which is false. - If
x = 1, then-1 < 1is true, so the rest is1 < -1which is false. - If
x = -1, then-2 < -1is true, so the rest is1 < -2, which is false. - If
x = 1, then-2 < 1is true, so the rest is1 < -2which is false.
answered Nov 15 '18 at 13:15
T.J. CrowderT.J. Crowder
694k12212381330
add a comment |
Because JavaScript allows implicit type coercion, in this case from boolean to number. The -1 < x results in a boolean, which is then implicitly coerced to a number (true = 1, false = 0) for the (result) < 1 part. So:
When
-1 < xis false, the second part is0 < 1which is true.When
-1 < xis true, the second part is1 < 1which is false.
This is covered in the abstract relational comparison algorithm in the spec, and the various operations it links to.
-1 < x < -1 is always false
-2 < x < 2 is always true
In the last 2 cases it seems it is just comparing the 2 ends of the expressions. How are those expressions evalued?
Using x = -1 and x = 1:
- If
x = -1, then-1 < xis false, so the rest is0 < -1, which is false. - If
x = 1, then-1 < 1is true, so the rest is1 < -1which is false. - If
x = -1, then-2 < -1is true, so the rest is1 < -2, which is false. - If
x = 1, then-2 < 1is true, so the rest is1 < -2which is false.
answered Nov 15 '18 at 13:15
T.J. CrowderT.J. Crowder
694k12212381330
add a comment |
Because JavaScript allows implicit type coercion, in this case from boolean to number. The -1 < x results in a boolean, which is then implicitly coerced to a number (true = 1, false = 0) for the (result) < 1 part. So:
When
-1 < xis false, the second part is0 < 1which is true.When
-1 < xis true, the second part is1 < 1which is false.
This is covered in the abstract relational comparison algorithm in the spec, and the various operations it links to.
-1 < x < -1 is always false
-2 < x < 2 is always true
In the last 2 cases it seems it is just comparing the 2 ends of the expressions. How are those expressions evalued?
Using x = -1 and x = 1:
- If
x = -1, then-1 < xis false, so the rest is0 < -1, which is false. - If
x = 1, then-1 < 1is true, so the rest is1 < -1which is false. - If
x = -1, then-2 < -1is true, so the rest is1 < -2, which is false. - If
x = 1, then-2 < 1is true, so the rest is1 < -2which is false.
answered Nov 15 '18 at 13:15
T.J. CrowderT.J. Crowder
694k12212381330
Because JavaScript allows implicit type coercion, in this case from boolean to number. The -1 < x results in a boolean, which is then implicitly coerced to a number (true = 1, false = 0) for the (result) < 1 part. So:
When
-1 < xis false, the second part is0 < 1which is true.When
-1 < xis true, the second part is1 < 1which is false.
This is covered in the abstract relational comparison algorithm in the spec, and the various operations it links to.
-1 < x < -1 is always false
-2 < x < 2 is always true
In the last 2 cases it seems it is just comparing the 2 ends of the expressions. How are those expressions evalued?
Using x = -1 and x = 1:
- If
x = -1, then-1 < xis false, so the rest is0 < -1, which is false. - If
x = 1, then-1 < 1is true, so the rest is1 < -1which is false. - If
x = -1, then-2 < -1is true, so the rest is1 < -2, which is false. - If
x = 1, then-2 < 1is true, so the rest is1 < -2which is false.
answered Nov 15 '18 at 13:15
T.J. CrowderT.J. Crowder
694k12212381330
edited Nov 15 '18 at 13:20
answered Nov 15 '18 at 13:15
T.J. CrowderT.J. Crowder
694k12212381330
answered Nov 15 '18 at 13:15
T.J. CrowderT.J. Crowder
694k12212381330
answered Nov 15 '18 at 13:15
T.J. CrowderT.J. Crowder
694k12212381330
694k12212381330
add a comment |
add a comment |
I think that this happens because javascript implicitly considers true to have the value 1 and false to have the value 0.
When you do -1 < x < 1, what you're actually doing is (-1 < x) < 1, or true < 1 if x = 0 which is false.
However, if x=-2, (-1 < x < -1) will return true as false < 1 is true. Hope this helps.
You can read more about this here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
answered Nov 15 '18 at 13:20
theapologisttheapologist
576215
add a comment |
I think that this happens because javascript implicitly considers true to have the value 1 and false to have the value 0.
When you do -1 < x < 1, what you're actually doing is (-1 < x) < 1, or true < 1 if x = 0 which is false.
However, if x=-2, (-1 < x < -1) will return true as false < 1 is true. Hope this helps.
You can read more about this here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
answered Nov 15 '18 at 13:20
theapologisttheapologist
576215
add a comment |
I think that this happens because javascript implicitly considers true to have the value 1 and false to have the value 0.
When you do -1 < x < 1, what you're actually doing is (-1 < x) < 1, or true < 1 if x = 0 which is false.
However, if x=-2, (-1 < x < -1) will return true as false < 1 is true. Hope this helps.
You can read more about this here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
answered Nov 15 '18 at 13:20
theapologisttheapologist
576215
I think that this happens because javascript implicitly considers true to have the value 1 and false to have the value 0.
When you do -1 < x < 1, what you're actually doing is (-1 < x) < 1, or true < 1 if x = 0 which is false.
However, if x=-2, (-1 < x < -1) will return true as false < 1 is true. Hope this helps.
You can read more about this here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
answered Nov 15 '18 at 13:20
theapologisttheapologist
576215
edited Nov 15 '18 at 13:25
answered Nov 15 '18 at 13:20
theapologisttheapologist
576215
answered Nov 15 '18 at 13:20
theapologisttheapologist
576215
answered Nov 15 '18 at 13:20
theapologisttheapologist
576215
576215
add a comment |
add a comment |
The reason is that JS interprets your expression.
if((-1 < x) < 1)
console.log('x: ', x)
Trying using braces...
answered Nov 15 '18 at 13:20
Ram VenkatRam Venkat
787
add a comment |
The reason is that JS interprets your expression.
if((-1 < x) < 1)
console.log('x: ', x)
Trying using braces...
answered Nov 15 '18 at 13:20
Ram VenkatRam Venkat
787
add a comment |
The reason is that JS interprets your expression.
if((-1 < x) < 1)
console.log('x: ', x)
Trying using braces...
answered Nov 15 '18 at 13:20
Ram VenkatRam Venkat
787
The reason is that JS interprets your expression.
if((-1 < x) < 1)
console.log('x: ', x)
Trying using braces...
answered Nov 15 '18 at 13:20
Ram VenkatRam Venkat
787
answered Nov 15 '18 at 13:20
Ram VenkatRam Venkat
787
answered Nov 15 '18 at 13:20
Ram VenkatRam Venkat
787
answered Nov 15 '18 at 13:20
Ram VenkatRam Venkat
787
787
add a comment |
add a comment |
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1
They are evaluated as
(-1 < x) < 1, which is equivalent to(-1 < x) != trueor!(-1 < x)– Bergi
Nov 15 '18 at 13:15
What you really need is
-1 < x && x < 1– yunzen
Nov 15 '18 at 13:24
@HerrSerker I know how to do proper comparisons, I was just asking how JS evaluated what in other languages is called interval comparisons.
– Andrea Bergonzo
Nov 15 '18 at 13:59
I knew this. But it would be interesting to other people who want to know, how it's done right
– yunzen
Nov 15 '18 at 14:34