How is “for(const [[[[[a, b, c]]]]] in [0, 0]);” even valid?
up vote
22
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Writing dumb code, I've just found something weird.
for(const [[[[[fancy, loop]]]]] in [0, 0]) {
console.log(fancy, loop);
}// Chrome 70.0.3538.77 says:
// 0 undefined
// 1 undefined
It's like assigning 0 and 1 to [[[[[fancy, loop]]]]], which, is array destructurings taking place and supposed to throw an error, isn't it? Or not. It's just my thought getting me confused right now.
Could you please tell me how it is valid and works with no error? What am I missing?
javascript
asked Nov 5 at 0:20
K._
2,29521436
add a comment |
up vote
22
down vote
favorite
Writing dumb code, I've just found something weird.
for(const [[[[[fancy, loop]]]]] in [0, 0]) {
console.log(fancy, loop);
}// Chrome 70.0.3538.77 says:
// 0 undefined
// 1 undefined
It's like assigning 0 and 1 to [[[[[fancy, loop]]]]], which, is array destructurings taking place and supposed to throw an error, isn't it? Or not. It's just my thought getting me confused right now.
Could you please tell me how it is valid and works with no error? What am I missing?
javascript
asked Nov 5 at 0:20
K._
2,29521436
@RobG Yeah, but, how could it destructure0and1?
– K._
Nov 5 at 0:25
2
you can always throw the code into a transpiler to see what is happening
– Bravo
Nov 5 at 0:25
2
babeljs.io/repl/…
– Bravo
Nov 5 at 0:26
17
I guess you discovered a new question for javascript job interviews. ;-)
– RobG
Nov 5 at 0:49
4
^ @RobG for anyone wondering, this should be taken as sarcasm
– Pierre Arlaud
Nov 5 at 9:28
add a comment |
up vote
22
down vote
favorite
up vote
22
down vote
favorite
Writing dumb code, I've just found something weird.
for(const [[[[[fancy, loop]]]]] in [0, 0]) {
console.log(fancy, loop);
}// Chrome 70.0.3538.77 says:
// 0 undefined
// 1 undefined
It's like assigning 0 and 1 to [[[[[fancy, loop]]]]], which, is array destructurings taking place and supposed to throw an error, isn't it? Or not. It's just my thought getting me confused right now.
Could you please tell me how it is valid and works with no error? What am I missing?
javascript
asked Nov 5 at 0:20
K._
2,29521436
Writing dumb code, I've just found something weird.
for(const [[[[[fancy, loop]]]]] in [0, 0]) {
console.log(fancy, loop);
}// Chrome 70.0.3538.77 says:
// 0 undefined
// 1 undefined
It's like assigning 0 and 1 to [[[[[fancy, loop]]]]], which, is array destructurings taking place and supposed to throw an error, isn't it? Or not. It's just my thought getting me confused right now.
Could you please tell me how it is valid and works with no error? What am I missing?
for(const [[[[[fancy, loop]]]]] in [0, 0]) {
console.log(fancy, loop);
}for(const [[[[[fancy, loop]]]]] in [0, 0]) {
console.log(fancy, loop);
}javascript
javascript
asked Nov 5 at 0:20
K._
2,29521436
asked Nov 5 at 0:20
K._
2,29521436
edited Nov 5 at 0:23
asked Nov 5 at 0:20
K._
2,29521436
asked Nov 5 at 0:20
K._
2,29521436
asked Nov 5 at 0:20
K._
2,29521436
2,29521436
@RobG Yeah, but, how could it destructure0and1?
– K._
Nov 5 at 0:25
2
you can always throw the code into a transpiler to see what is happening
– Bravo
Nov 5 at 0:25
2
babeljs.io/repl/…
– Bravo
Nov 5 at 0:26
17
I guess you discovered a new question for javascript job interviews. ;-)
– RobG
Nov 5 at 0:49
4
^ @RobG for anyone wondering, this should be taken as sarcasm
– Pierre Arlaud
Nov 5 at 9:28
add a comment |
@RobG Yeah, but, how could it destructure0and1?
– K._
Nov 5 at 0:25
2
you can always throw the code into a transpiler to see what is happening
– Bravo
Nov 5 at 0:25
2
babeljs.io/repl/…
– Bravo
Nov 5 at 0:26
17
I guess you discovered a new question for javascript job interviews. ;-)
– RobG
Nov 5 at 0:49
4
^ @RobG for anyone wondering, this should be taken as sarcasm
– Pierre Arlaud
Nov 5 at 9:28
@RobG Yeah, but, how could it destructure
0 and 1?– K._
Nov 5 at 0:25
@RobG Yeah, but, how could it destructure
0 and 1?– K._
Nov 5 at 0:25
2
2
you can always throw the code into a transpiler to see what is happening
– Bravo
Nov 5 at 0:25
you can always throw the code into a transpiler to see what is happening
– Bravo
Nov 5 at 0:25
2
2
babeljs.io/repl/…
– Bravo
Nov 5 at 0:26
babeljs.io/repl/…
– Bravo
Nov 5 at 0:26
17
17
I guess you discovered a new question for javascript job interviews. ;-)
– RobG
Nov 5 at 0:49
I guess you discovered a new question for javascript job interviews. ;-)
– RobG
Nov 5 at 0:49
4
4
^ @RobG for anyone wondering, this should be taken as sarcasm
– Pierre Arlaud
Nov 5 at 9:28
^ @RobG for anyone wondering, this should be taken as sarcasm
– Pierre Arlaud
Nov 5 at 9:28
add a comment |
2 Answers
2
active
oldest
votes
up vote
21
down vote
accepted
It's not assigning 0 and 1 to [[[[[fancy, loop]]]]]. You're looping over the keys of [0, 0], because you used in instead of of, and those keys are strings.
The string "0" is an iterable whose sole element is "0". Assigning "0" to [[[[[fancy, loop]]]]] repeatedly unpacks "0" and gets "0", until eventually it gets down to
[fancy, loop] = "0"
at which point the final unpacking assigns "0" to fancy and undefined to loop.
answered Nov 5 at 0:39
user2357112
146k12149233
1
This doesn't fully explain what is occurring though, there is a difference between[fancy, loop]and[[fancy, loop]]given[,,,,,,,,,0,0](i.e. try it with property names of more than one character). The assignments are different so it doesn't simply reduce to[fancy, loop].
– RobG
Nov 5 at 0:54
1
@RobG: You skipped the first four layers of unpacking, each of which only takes one element."10"gets unpacked into"1"and"0", and the"0"gets discarded in the first layer of the destructuring assignment. By the time we hit the[fancy, loop]unpacking, the"0"has been gone for the last three layers.
– user2357112
Nov 5 at 1:12
Yes, that's the missing part. ;-)
– RobG
Nov 5 at 1:13
add a comment |
up vote
4
down vote
You're using in instead of of so you're getting the properties of the array not the values. For this case you're getting the array indexes as strings (0, 1). You're basically destructuring a string with length of 1 every time. So you always get the first character of every iterated property
answered Nov 5 at 0:40
lleon
716167
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
21
down vote
accepted
It's not assigning 0 and 1 to [[[[[fancy, loop]]]]]. You're looping over the keys of [0, 0], because you used in instead of of, and those keys are strings.
The string "0" is an iterable whose sole element is "0". Assigning "0" to [[[[[fancy, loop]]]]] repeatedly unpacks "0" and gets "0", until eventually it gets down to
[fancy, loop] = "0"
at which point the final unpacking assigns "0" to fancy and undefined to loop.
answered Nov 5 at 0:39
user2357112
146k12149233
1
This doesn't fully explain what is occurring though, there is a difference between[fancy, loop]and[[fancy, loop]]given[,,,,,,,,,0,0](i.e. try it with property names of more than one character). The assignments are different so it doesn't simply reduce to[fancy, loop].
– RobG
Nov 5 at 0:54
1
@RobG: You skipped the first four layers of unpacking, each of which only takes one element."10"gets unpacked into"1"and"0", and the"0"gets discarded in the first layer of the destructuring assignment. By the time we hit the[fancy, loop]unpacking, the"0"has been gone for the last three layers.
– user2357112
Nov 5 at 1:12
Yes, that's the missing part. ;-)
– RobG
Nov 5 at 1:13
add a comment |
up vote
21
down vote
accepted
It's not assigning 0 and 1 to [[[[[fancy, loop]]]]]. You're looping over the keys of [0, 0], because you used in instead of of, and those keys are strings.
The string "0" is an iterable whose sole element is "0". Assigning "0" to [[[[[fancy, loop]]]]] repeatedly unpacks "0" and gets "0", until eventually it gets down to
[fancy, loop] = "0"
at which point the final unpacking assigns "0" to fancy and undefined to loop.
answered Nov 5 at 0:39
user2357112
146k12149233
1
This doesn't fully explain what is occurring though, there is a difference between[fancy, loop]and[[fancy, loop]]given[,,,,,,,,,0,0](i.e. try it with property names of more than one character). The assignments are different so it doesn't simply reduce to[fancy, loop].
– RobG
Nov 5 at 0:54
1
@RobG: You skipped the first four layers of unpacking, each of which only takes one element."10"gets unpacked into"1"and"0", and the"0"gets discarded in the first layer of the destructuring assignment. By the time we hit the[fancy, loop]unpacking, the"0"has been gone for the last three layers.
– user2357112
Nov 5 at 1:12
Yes, that's the missing part. ;-)
– RobG
Nov 5 at 1:13
add a comment |
up vote
21
down vote
accepted
up vote
21
down vote
accepted
It's not assigning 0 and 1 to [[[[[fancy, loop]]]]]. You're looping over the keys of [0, 0], because you used in instead of of, and those keys are strings.
The string "0" is an iterable whose sole element is "0". Assigning "0" to [[[[[fancy, loop]]]]] repeatedly unpacks "0" and gets "0", until eventually it gets down to
[fancy, loop] = "0"
at which point the final unpacking assigns "0" to fancy and undefined to loop.
answered Nov 5 at 0:39
user2357112
146k12149233
It's not assigning 0 and 1 to [[[[[fancy, loop]]]]]. You're looping over the keys of [0, 0], because you used in instead of of, and those keys are strings.
The string "0" is an iterable whose sole element is "0". Assigning "0" to [[[[[fancy, loop]]]]] repeatedly unpacks "0" and gets "0", until eventually it gets down to
[fancy, loop] = "0"
at which point the final unpacking assigns "0" to fancy and undefined to loop.
answered Nov 5 at 0:39
user2357112
146k12149233
answered Nov 5 at 0:39
user2357112
146k12149233
answered Nov 5 at 0:39
user2357112
146k12149233
answered Nov 5 at 0:39
user2357112
146k12149233
146k12149233
1
This doesn't fully explain what is occurring though, there is a difference between[fancy, loop]and[[fancy, loop]]given[,,,,,,,,,0,0](i.e. try it with property names of more than one character). The assignments are different so it doesn't simply reduce to[fancy, loop].
– RobG
Nov 5 at 0:54
1
@RobG: You skipped the first four layers of unpacking, each of which only takes one element."10"gets unpacked into"1"and"0", and the"0"gets discarded in the first layer of the destructuring assignment. By the time we hit the[fancy, loop]unpacking, the"0"has been gone for the last three layers.
– user2357112
Nov 5 at 1:12
Yes, that's the missing part. ;-)
– RobG
Nov 5 at 1:13
add a comment |
1
This doesn't fully explain what is occurring though, there is a difference between[fancy, loop]and[[fancy, loop]]given[,,,,,,,,,0,0](i.e. try it with property names of more than one character). The assignments are different so it doesn't simply reduce to[fancy, loop].
– RobG
Nov 5 at 0:54
1
@RobG: You skipped the first four layers of unpacking, each of which only takes one element."10"gets unpacked into"1"and"0", and the"0"gets discarded in the first layer of the destructuring assignment. By the time we hit the[fancy, loop]unpacking, the"0"has been gone for the last three layers.
– user2357112
Nov 5 at 1:12
Yes, that's the missing part. ;-)
– RobG
Nov 5 at 1:13
1
1
This doesn't fully explain what is occurring though, there is a difference between
[fancy, loop] and [[fancy, loop]] given [,,,,,,,,,0,0] (i.e. try it with property names of more than one character). The assignments are different so it doesn't simply reduce to [fancy, loop].– RobG
Nov 5 at 0:54
This doesn't fully explain what is occurring though, there is a difference between
[fancy, loop] and [[fancy, loop]] given [,,,,,,,,,0,0] (i.e. try it with property names of more than one character). The assignments are different so it doesn't simply reduce to [fancy, loop].– RobG
Nov 5 at 0:54
1
1
@RobG: You skipped the first four layers of unpacking, each of which only takes one element.
"10" gets unpacked into "1" and "0", and the "0" gets discarded in the first layer of the destructuring assignment. By the time we hit the [fancy, loop] unpacking, the "0" has been gone for the last three layers.– user2357112
Nov 5 at 1:12
@RobG: You skipped the first four layers of unpacking, each of which only takes one element.
"10" gets unpacked into "1" and "0", and the "0" gets discarded in the first layer of the destructuring assignment. By the time we hit the [fancy, loop] unpacking, the "0" has been gone for the last three layers.– user2357112
Nov 5 at 1:12
Yes, that's the missing part. ;-)
– RobG
Nov 5 at 1:13
Yes, that's the missing part. ;-)
– RobG
Nov 5 at 1:13
add a comment |
up vote
4
down vote
You're using in instead of of so you're getting the properties of the array not the values. For this case you're getting the array indexes as strings (0, 1). You're basically destructuring a string with length of 1 every time. So you always get the first character of every iterated property
answered Nov 5 at 0:40
lleon
716167
add a comment |
up vote
4
down vote
You're using in instead of of so you're getting the properties of the array not the values. For this case you're getting the array indexes as strings (0, 1). You're basically destructuring a string with length of 1 every time. So you always get the first character of every iterated property
answered Nov 5 at 0:40
lleon
716167
add a comment |
up vote
4
down vote
up vote
4
down vote
You're using in instead of of so you're getting the properties of the array not the values. For this case you're getting the array indexes as strings (0, 1). You're basically destructuring a string with length of 1 every time. So you always get the first character of every iterated property
answered Nov 5 at 0:40
lleon
716167
You're using in instead of of so you're getting the properties of the array not the values. For this case you're getting the array indexes as strings (0, 1). You're basically destructuring a string with length of 1 every time. So you always get the first character of every iterated property
answered Nov 5 at 0:40
lleon
716167
edited Nov 5 at 10:43
answered Nov 5 at 0:40
lleon
716167
answered Nov 5 at 0:40
lleon
716167
answered Nov 5 at 0:40
lleon
716167
716167
add a comment |
add a comment |
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@RobG Yeah, but, how could it destructure
0and1?– K._
Nov 5 at 0:25
2
you can always throw the code into a transpiler to see what is happening
– Bravo
Nov 5 at 0:25
2
babeljs.io/repl/…
– Bravo
Nov 5 at 0:26
17
I guess you discovered a new question for javascript job interviews. ;-)
– RobG
Nov 5 at 0:49
4
^ @RobG for anyone wondering, this should be taken as sarcasm
– Pierre Arlaud
Nov 5 at 9:28