What is the purpose of spread syntax in this example? [duplicate]
This question already has an answer here:
Why does spread syntax convert my string into an array?
5 answers
Here,we created Hashmap with keys and value.Then we use array.map to get our result. i.e If we input "cat",we will get output as "dog".I didnt get how spread syntax is used.
var rule =
{
"c": "d",
"a": "o",
"t": "g",
"h": "a",
"e": "n",
"n": "t",
}
function output(str) {
return [...str].map(d => rule[d]).join('')
}
console.log(output('cat'))
javascript arrays
marked as duplicate by Mamun
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Nov 25 '18 at 9:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Why does spread syntax convert my string into an array?
5 answers
Here,we created Hashmap with keys and value.Then we use array.map to get our result. i.e If we input "cat",we will get output as "dog".I didnt get how spread syntax is used.
var rule =
{
"c": "d",
"a": "o",
"t": "g",
"h": "a",
"e": "n",
"n": "t",
}
function output(str) {
return [...str].map(d => rule[d]).join('')
}
console.log(output('cat'))
javascript arrays
marked as duplicate by Mamun
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Nov 25 '18 at 9:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Could you not have tried it? Just run[...'cat']
in the console.
– jonrsharpe
Nov 25 '18 at 9:39
add a comment |
This question already has an answer here:
Why does spread syntax convert my string into an array?
5 answers
Here,we created Hashmap with keys and value.Then we use array.map to get our result. i.e If we input "cat",we will get output as "dog".I didnt get how spread syntax is used.
var rule =
{
"c": "d",
"a": "o",
"t": "g",
"h": "a",
"e": "n",
"n": "t",
}
function output(str) {
return [...str].map(d => rule[d]).join('')
}
console.log(output('cat'))
javascript arrays
This question already has an answer here:
Why does spread syntax convert my string into an array?
5 answers
Here,we created Hashmap with keys and value.Then we use array.map to get our result. i.e If we input "cat",we will get output as "dog".I didnt get how spread syntax is used.
var rule =
{
"c": "d",
"a": "o",
"t": "g",
"h": "a",
"e": "n",
"n": "t",
}
function output(str) {
return [...str].map(d => rule[d]).join('')
}
console.log(output('cat'))
This question already has an answer here:
Why does spread syntax convert my string into an array?
5 answers
javascript arrays
javascript arrays
asked Nov 25 '18 at 9:32
Suman BasnetSuman Basnet
426
426
marked as duplicate by Mamun
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Nov 25 '18 at 9:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Mamun
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Nov 25 '18 at 9:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Could you not have tried it? Just run[...'cat']
in the console.
– jonrsharpe
Nov 25 '18 at 9:39
add a comment |
Could you not have tried it? Just run[...'cat']
in the console.
– jonrsharpe
Nov 25 '18 at 9:39
Could you not have tried it? Just run
[...'cat']
in the console.– jonrsharpe
Nov 25 '18 at 9:39
Could you not have tried it? Just run
[...'cat']
in the console.– jonrsharpe
Nov 25 '18 at 9:39
add a comment |
5 Answers
5
active
oldest
votes
Break it down:
[...str]
Gives: ['c', 'a', 't'];
In this case, the spread syntax breaks up the string into an array. Where each index represents a letter from the string.
.map(d => rule[d])
Converts the above array to:
['d', 'o', 'g']
As it goes through each value and uses the javascript object to get its value: (so 'c' maps to 'd' as rule['c'] = 'd'
, 'a' maps to 'o', etc...)
.join('')
Converts the above array to a string dog
Thus, doing:
console.log(output('cat'))
Gives: "dog"
add a comment |
spread syntax on a string turns it into an array consisting of its characters. This works because strings are iterable collections. http://es6-features.org/#SpreadOperator
add a comment |
In your case the spread operator spreads the string "cat" into an array of characters containing ['c','a','t']
Then it iterates over this array to provide the output string matches the values. There you get the string "dog" as output!
add a comment |
You would not be able to apply map
to the string directly, as String
does not have such a method. With the spread syntax you get an array of individual characters, and then map
can be applied to that array.
It is in fact better practice to use Array.from
instead, because it does not create the intermediate array and still allows you to execute a function on each character (using the second argument):
Array.from(str, d => rule[d]).join('')
add a comment |
It takes an iterable and returns each item of the iterable by using the spread syntax ...
.
In this case, a string is splitted to an array of characters.
['c', 'a', 't']
The same result could be achieved, if you use Array.from
, which has a built-in mapping parameter.
var rule = { c: "d", a: "o", t: "g", h: "a", e: "n", n: "t" }
function output(str) {
return Array.from(str, d => rule[d]).join('');
}
console.log(output('cat')) ;
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Break it down:
[...str]
Gives: ['c', 'a', 't'];
In this case, the spread syntax breaks up the string into an array. Where each index represents a letter from the string.
.map(d => rule[d])
Converts the above array to:
['d', 'o', 'g']
As it goes through each value and uses the javascript object to get its value: (so 'c' maps to 'd' as rule['c'] = 'd'
, 'a' maps to 'o', etc...)
.join('')
Converts the above array to a string dog
Thus, doing:
console.log(output('cat'))
Gives: "dog"
add a comment |
Break it down:
[...str]
Gives: ['c', 'a', 't'];
In this case, the spread syntax breaks up the string into an array. Where each index represents a letter from the string.
.map(d => rule[d])
Converts the above array to:
['d', 'o', 'g']
As it goes through each value and uses the javascript object to get its value: (so 'c' maps to 'd' as rule['c'] = 'd'
, 'a' maps to 'o', etc...)
.join('')
Converts the above array to a string dog
Thus, doing:
console.log(output('cat'))
Gives: "dog"
add a comment |
Break it down:
[...str]
Gives: ['c', 'a', 't'];
In this case, the spread syntax breaks up the string into an array. Where each index represents a letter from the string.
.map(d => rule[d])
Converts the above array to:
['d', 'o', 'g']
As it goes through each value and uses the javascript object to get its value: (so 'c' maps to 'd' as rule['c'] = 'd'
, 'a' maps to 'o', etc...)
.join('')
Converts the above array to a string dog
Thus, doing:
console.log(output('cat'))
Gives: "dog"
Break it down:
[...str]
Gives: ['c', 'a', 't'];
In this case, the spread syntax breaks up the string into an array. Where each index represents a letter from the string.
.map(d => rule[d])
Converts the above array to:
['d', 'o', 'g']
As it goes through each value and uses the javascript object to get its value: (so 'c' maps to 'd' as rule['c'] = 'd'
, 'a' maps to 'o', etc...)
.join('')
Converts the above array to a string dog
Thus, doing:
console.log(output('cat'))
Gives: "dog"
edited Nov 28 '18 at 9:05
answered Nov 25 '18 at 9:40
Nick ParsonsNick Parsons
8,7282824
8,7282824
add a comment |
add a comment |
spread syntax on a string turns it into an array consisting of its characters. This works because strings are iterable collections. http://es6-features.org/#SpreadOperator
add a comment |
spread syntax on a string turns it into an array consisting of its characters. This works because strings are iterable collections. http://es6-features.org/#SpreadOperator
add a comment |
spread syntax on a string turns it into an array consisting of its characters. This works because strings are iterable collections. http://es6-features.org/#SpreadOperator
spread syntax on a string turns it into an array consisting of its characters. This works because strings are iterable collections. http://es6-features.org/#SpreadOperator
answered Nov 25 '18 at 9:36
lluisrojasslluisrojass
11119
11119
add a comment |
add a comment |
In your case the spread operator spreads the string "cat" into an array of characters containing ['c','a','t']
Then it iterates over this array to provide the output string matches the values. There you get the string "dog" as output!
add a comment |
In your case the spread operator spreads the string "cat" into an array of characters containing ['c','a','t']
Then it iterates over this array to provide the output string matches the values. There you get the string "dog" as output!
add a comment |
In your case the spread operator spreads the string "cat" into an array of characters containing ['c','a','t']
Then it iterates over this array to provide the output string matches the values. There you get the string "dog" as output!
In your case the spread operator spreads the string "cat" into an array of characters containing ['c','a','t']
Then it iterates over this array to provide the output string matches the values. There you get the string "dog" as output!
answered Nov 25 '18 at 9:38
LuciferLucifer
304211
304211
add a comment |
add a comment |
You would not be able to apply map
to the string directly, as String
does not have such a method. With the spread syntax you get an array of individual characters, and then map
can be applied to that array.
It is in fact better practice to use Array.from
instead, because it does not create the intermediate array and still allows you to execute a function on each character (using the second argument):
Array.from(str, d => rule[d]).join('')
add a comment |
You would not be able to apply map
to the string directly, as String
does not have such a method. With the spread syntax you get an array of individual characters, and then map
can be applied to that array.
It is in fact better practice to use Array.from
instead, because it does not create the intermediate array and still allows you to execute a function on each character (using the second argument):
Array.from(str, d => rule[d]).join('')
add a comment |
You would not be able to apply map
to the string directly, as String
does not have such a method. With the spread syntax you get an array of individual characters, and then map
can be applied to that array.
It is in fact better practice to use Array.from
instead, because it does not create the intermediate array and still allows you to execute a function on each character (using the second argument):
Array.from(str, d => rule[d]).join('')
You would not be able to apply map
to the string directly, as String
does not have such a method. With the spread syntax you get an array of individual characters, and then map
can be applied to that array.
It is in fact better practice to use Array.from
instead, because it does not create the intermediate array and still allows you to execute a function on each character (using the second argument):
Array.from(str, d => rule[d]).join('')
answered Nov 25 '18 at 9:39
trincottrincot
125k1588121
125k1588121
add a comment |
add a comment |
It takes an iterable and returns each item of the iterable by using the spread syntax ...
.
In this case, a string is splitted to an array of characters.
['c', 'a', 't']
The same result could be achieved, if you use Array.from
, which has a built-in mapping parameter.
var rule = { c: "d", a: "o", t: "g", h: "a", e: "n", n: "t" }
function output(str) {
return Array.from(str, d => rule[d]).join('');
}
console.log(output('cat')) ;
add a comment |
It takes an iterable and returns each item of the iterable by using the spread syntax ...
.
In this case, a string is splitted to an array of characters.
['c', 'a', 't']
The same result could be achieved, if you use Array.from
, which has a built-in mapping parameter.
var rule = { c: "d", a: "o", t: "g", h: "a", e: "n", n: "t" }
function output(str) {
return Array.from(str, d => rule[d]).join('');
}
console.log(output('cat')) ;
add a comment |
It takes an iterable and returns each item of the iterable by using the spread syntax ...
.
In this case, a string is splitted to an array of characters.
['c', 'a', 't']
The same result could be achieved, if you use Array.from
, which has a built-in mapping parameter.
var rule = { c: "d", a: "o", t: "g", h: "a", e: "n", n: "t" }
function output(str) {
return Array.from(str, d => rule[d]).join('');
}
console.log(output('cat')) ;
It takes an iterable and returns each item of the iterable by using the spread syntax ...
.
In this case, a string is splitted to an array of characters.
['c', 'a', 't']
The same result could be achieved, if you use Array.from
, which has a built-in mapping parameter.
var rule = { c: "d", a: "o", t: "g", h: "a", e: "n", n: "t" }
function output(str) {
return Array.from(str, d => rule[d]).join('');
}
console.log(output('cat')) ;
var rule = { c: "d", a: "o", t: "g", h: "a", e: "n", n: "t" }
function output(str) {
return Array.from(str, d => rule[d]).join('');
}
console.log(output('cat')) ;
var rule = { c: "d", a: "o", t: "g", h: "a", e: "n", n: "t" }
function output(str) {
return Array.from(str, d => rule[d]).join('');
}
console.log(output('cat')) ;
edited Nov 28 '18 at 9:07
Nick Parsons
8,7282824
8,7282824
answered Nov 25 '18 at 9:38
Nina ScholzNina Scholz
187k1596172
187k1596172
add a comment |
add a comment |
Could you not have tried it? Just run
[...'cat']
in the console.– jonrsharpe
Nov 25 '18 at 9:39