How nested function access variables in Javascript at a later time without declaration?
In below code how the parameters passed to 'DPerson' function survive even after call to this function has ended successfully. If this is related to how stack and heaps work, can anyone explain it in more detail with respect to this example.
I was of the understanding that, a 'var name = "someValue"' property need to be created in at least one of the execution contexts to print something in console log. But apparently my understanding is wrong.
function DPerson(name, age, job) {
var o = new Object();
o.sayName = function() {
console.log(name);
}
return o;
}
var dperson1 = new DPerson("Ahu", 55, "Wild life expert");
dperson1.sayName();javascript
edited Nov 16 '18 at 2:20
Olian04
2,18511136
asked Nov 15 '18 at 23:23
MohitMohit
15318
add a comment |
In below code how the parameters passed to 'DPerson' function survive even after call to this function has ended successfully. If this is related to how stack and heaps work, can anyone explain it in more detail with respect to this example.
I was of the understanding that, a 'var name = "someValue"' property need to be created in at least one of the execution contexts to print something in console log. But apparently my understanding is wrong.
function DPerson(name, age, job) {
var o = new Object();
o.sayName = function() {
console.log(name);
}
return o;
}
var dperson1 = new DPerson("Ahu", 55, "Wild life expert");
dperson1.sayName();javascript
edited Nov 16 '18 at 2:20
Olian04
2,18511136
asked Nov 15 '18 at 23:23
MohitMohit
15318
The concept you're looking for is called a closure :)
– Olian04
Nov 15 '18 at 23:32
1
In this particular code onlynamewill "survive" - notageorjob. And the reason it does survive is because thesayNamemethod of the returned object has a "closure" over thenamevariable. There are probably thousands of good explanations online of how closure works in Javascript, I suggest you read some if you're confused by this.
– Robin Zigmond
Nov 15 '18 at 23:33
add a comment |
In below code how the parameters passed to 'DPerson' function survive even after call to this function has ended successfully. If this is related to how stack and heaps work, can anyone explain it in more detail with respect to this example.
I was of the understanding that, a 'var name = "someValue"' property need to be created in at least one of the execution contexts to print something in console log. But apparently my understanding is wrong.
function DPerson(name, age, job) {
var o = new Object();
o.sayName = function() {
console.log(name);
}
return o;
}
var dperson1 = new DPerson("Ahu", 55, "Wild life expert");
dperson1.sayName();javascript
edited Nov 16 '18 at 2:20
Olian04
2,18511136
asked Nov 15 '18 at 23:23
MohitMohit
15318
In below code how the parameters passed to 'DPerson' function survive even after call to this function has ended successfully. If this is related to how stack and heaps work, can anyone explain it in more detail with respect to this example.
I was of the understanding that, a 'var name = "someValue"' property need to be created in at least one of the execution contexts to print something in console log. But apparently my understanding is wrong.
function DPerson(name, age, job) {
var o = new Object();
o.sayName = function() {
console.log(name);
}
return o;
}
var dperson1 = new DPerson("Ahu", 55, "Wild life expert");
dperson1.sayName();function DPerson(name, age, job) {
var o = new Object();
o.sayName = function() {
console.log(name);
}
return o;
}
var dperson1 = new DPerson("Ahu", 55, "Wild life expert");
dperson1.sayName();function DPerson(name, age, job) {
var o = new Object();
o.sayName = function() {
console.log(name);
}
return o;
}
var dperson1 = new DPerson("Ahu", 55, "Wild life expert");
dperson1.sayName();javascript
javascript
edited Nov 16 '18 at 2:20
Olian04
2,18511136
asked Nov 15 '18 at 23:23
MohitMohit
15318
edited Nov 16 '18 at 2:20
Olian04
2,18511136
asked Nov 15 '18 at 23:23
MohitMohit
15318
edited Nov 16 '18 at 2:20
Olian04
2,18511136
edited Nov 16 '18 at 2:20
Olian04
2,18511136
edited Nov 16 '18 at 2:20
Olian04
2,18511136
2,18511136
asked Nov 15 '18 at 23:23
MohitMohit
15318
asked Nov 15 '18 at 23:23
MohitMohit
15318
asked Nov 15 '18 at 23:23
MohitMohit
15318
15318
The concept you're looking for is called a closure :)
– Olian04
Nov 15 '18 at 23:32
1
In this particular code onlynamewill "survive" - notageorjob. And the reason it does survive is because thesayNamemethod of the returned object has a "closure" over thenamevariable. There are probably thousands of good explanations online of how closure works in Javascript, I suggest you read some if you're confused by this.
– Robin Zigmond
Nov 15 '18 at 23:33
add a comment |
The concept you're looking for is called a closure :)
– Olian04
Nov 15 '18 at 23:32
1
In this particular code onlynamewill "survive" - notageorjob. And the reason it does survive is because thesayNamemethod of the returned object has a "closure" over thenamevariable. There are probably thousands of good explanations online of how closure works in Javascript, I suggest you read some if you're confused by this.
– Robin Zigmond
Nov 15 '18 at 23:33
The concept you're looking for is called a closure :)
– Olian04
Nov 15 '18 at 23:32
The concept you're looking for is called a closure :)
– Olian04
Nov 15 '18 at 23:32
1
1
In this particular code only
name will "survive" - not age or job. And the reason it does survive is because the sayName method of the returned object has a "closure" over the name variable. There are probably thousands of good explanations online of how closure works in Javascript, I suggest you read some if you're confused by this.– Robin Zigmond
Nov 15 '18 at 23:33
In this particular code only
name will "survive" - not age or job. And the reason it does survive is because the sayName method of the returned object has a "closure" over the name variable. There are probably thousands of good explanations online of how closure works in Javascript, I suggest you read some if you're confused by this.– Robin Zigmond
Nov 15 '18 at 23:33
add a comment |
2 Answers
2
active
oldest
votes
What you are seeing here is a closure. When you define a function inside another function, the child function has access to the lexical environment of the parent and keeps that even after the parent has returned. name is defined in the parent when you create the parameters to DPerson. The child function, o.sayName has access to this and retains access to it after the parent returns.
See MDN - closures for more
answered Nov 15 '18 at 23:32
Mark MeyerMark Meyer
39.5k33162
add a comment |
Yes variables in javascript survive even after the function ended. A function is in short a scope in javascript.
answered Nov 15 '18 at 23:37
Edwin Dijas ChiwonaEdwin Dijas Chiwona
35129
add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
oldest
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What you are seeing here is a closure. When you define a function inside another function, the child function has access to the lexical environment of the parent and keeps that even after the parent has returned. name is defined in the parent when you create the parameters to DPerson. The child function, o.sayName has access to this and retains access to it after the parent returns.
See MDN - closures for more
answered Nov 15 '18 at 23:32
Mark MeyerMark Meyer
39.5k33162
add a comment |
What you are seeing here is a closure. When you define a function inside another function, the child function has access to the lexical environment of the parent and keeps that even after the parent has returned. name is defined in the parent when you create the parameters to DPerson. The child function, o.sayName has access to this and retains access to it after the parent returns.
See MDN - closures for more
answered Nov 15 '18 at 23:32
Mark MeyerMark Meyer
39.5k33162
add a comment |
What you are seeing here is a closure. When you define a function inside another function, the child function has access to the lexical environment of the parent and keeps that even after the parent has returned. name is defined in the parent when you create the parameters to DPerson. The child function, o.sayName has access to this and retains access to it after the parent returns.
See MDN - closures for more
answered Nov 15 '18 at 23:32
Mark MeyerMark Meyer
39.5k33162
What you are seeing here is a closure. When you define a function inside another function, the child function has access to the lexical environment of the parent and keeps that even after the parent has returned. name is defined in the parent when you create the parameters to DPerson. The child function, o.sayName has access to this and retains access to it after the parent returns.
See MDN - closures for more
answered Nov 15 '18 at 23:32
Mark MeyerMark Meyer
39.5k33162
answered Nov 15 '18 at 23:32
Mark MeyerMark Meyer
39.5k33162
answered Nov 15 '18 at 23:32
Mark MeyerMark Meyer
39.5k33162
answered Nov 15 '18 at 23:32
Mark MeyerMark Meyer
39.5k33162
39.5k33162
add a comment |
add a comment |
Yes variables in javascript survive even after the function ended. A function is in short a scope in javascript.
answered Nov 15 '18 at 23:37
Edwin Dijas ChiwonaEdwin Dijas Chiwona
35129
add a comment |
Yes variables in javascript survive even after the function ended. A function is in short a scope in javascript.
answered Nov 15 '18 at 23:37
Edwin Dijas ChiwonaEdwin Dijas Chiwona
35129
add a comment |
Yes variables in javascript survive even after the function ended. A function is in short a scope in javascript.
answered Nov 15 '18 at 23:37
Edwin Dijas ChiwonaEdwin Dijas Chiwona
35129
Yes variables in javascript survive even after the function ended. A function is in short a scope in javascript.
answered Nov 15 '18 at 23:37
Edwin Dijas ChiwonaEdwin Dijas Chiwona
35129
answered Nov 15 '18 at 23:37
Edwin Dijas ChiwonaEdwin Dijas Chiwona
35129
answered Nov 15 '18 at 23:37
Edwin Dijas ChiwonaEdwin Dijas Chiwona
35129
answered Nov 15 '18 at 23:37
Edwin Dijas ChiwonaEdwin Dijas Chiwona
35129
35129
add a comment |
add a comment |
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The concept you're looking for is called a closure :)
– Olian04
Nov 15 '18 at 23:32
1
In this particular code only
namewill "survive" - notageorjob. And the reason it does survive is because thesayNamemethod of the returned object has a "closure" over thenamevariable. There are probably thousands of good explanations online of how closure works in Javascript, I suggest you read some if you're confused by this.– Robin Zigmond
Nov 15 '18 at 23:33