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improve formatting and grammar

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edited Dec 26, 2017 at 22:46

YakovL

edited Dec 26, 2017 at 22:46

YakovL

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We can initialize i with let or initialize a new variable index with let and make it equal to i. So when the call is being made, index will be used and its scope will end after initialization. And for calling, index will be initialized again:
We can initialize i with let or initialize a new variable index with let and make it equal to i. So when the call is being made, index will be used and its scope will end after initialization. And for calling, index will be initialized again:
```
 var funcs = [];
 for (var i = 0; i < 3; i++) { 
 let index = i;
 funcs[i] = function() { 
 console.log("My value: " + index); 
 };
 }
 for (var j = 0; j < 3; j++) {
 funcs[j](); 
 }
```

Other Option can be to introduce a tempFunc which returns the actual function:

 var funcs = [];
 function tempFunc(i){
 return function(){
 console.log("My value: " + i);
 };
 }
 for (var i = 0; i < 3; i++) { 
 funcs[i] = tempFunc(i); 
 }
 for (var j = 0; j < 3; j++) {
 funcs[j](); 
 }

var funcs = [];
for (var i = 0; i < 3; i++) { 
 let index = i;
 funcs[i] = function() { 
 console.log("My value: " + index); 
 };
}
for (var j = 0; j < 3; j++) {
 funcs[j](); 
}

Other Option can be to introduce a tempFunc which returns the actual function:

var funcs = [];
function tempFunc(i){
 return function(){
 console.log("My value: " + i);
 };
}
for (var i = 0; i < 3; i++) { 
 funcs[i] = tempFunc(i); 
 
}
for (var j = 0; j < 3; j++) {
 funcs[j](); 
}

We can initialize i with let or initialize a new variable index with let and make it equal to i. So when the call is being made, index will be used and its scope will end after initialization. And for calling, index will be initialized again:

var funcs = [];
for (var i = 0; i < 3; i++) { 
 let index = i;
 funcs[i] = function() { 
 console.log("My value: " + index); 
 };
}
for (var j = 0; j < 3; j++) {
 funcs[j](); 
}

Other Option can be to introduce a tempFunc which returns the actual function:

var funcs = [];
function tempFunc(i){
 return function(){
 console.log("My value: " + i);
 };
}
for (var i = 0; i < 3; i++) { 
 funcs[i] = tempFunc(i); 
 
}
for (var j = 0; j < 3; j++) {
 funcs[j](); 
}

We can initialize i with let or initialize a new variable index with let and make it equal to i. So when the call is being made, index will be used and its scope will end after initialization. And for calling, index will be initialized again:
```
 var funcs = [];
 for (var i = 0; i < 3; i++) { 
 let index = i;
 funcs[i] = function() { 
 console.log("My value: " + index); 
 };
 }
 for (var j = 0; j < 3; j++) {
 funcs[j](); 
 }
```

Other Option can be to introduce a tempFunc which returns the actual function:

 var funcs = [];
 function tempFunc(i){
 return function(){
 console.log("My value: " + i);
 };
 }
 for (var i = 0; i < 3; i++) { 
 funcs[i] = tempFunc(i); 
 }
 for (var j = 0; j < 3; j++) {
 funcs[j](); 
 }

formatting and grammatical improvement(s)

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edit approved Dec 26, 2017 at 22:46

Nae

edit approved Dec 26, 2017 at 22:46

Nae

15.5k
8
62
89

First of all, understand whatswhat's wrong with this code.:

hereHere when the funcs[] array is being initialized, i is being incremented, the funcsfuncs array is initialized and the size of func array becomes 3, so i = 3,. Now when the funcs[j]() is called, it is again using the variable variable i, which has already been incremented to 3.

Now to solve this, we have many options. Below are two of them.:

We can initialize i with let or initialize a new variable index with let and make it equal to i. soSo when the call is being made, index will be used and its scope will end after initialization. And for calling, index will be initialized again.:

Other Option can be to introduce a tempFunc which returns the actual function.:

First of all, understand whats wrong with this code.

here when the funcs[] array is being initialized, i is being incremented, the funcs array is initialized and the size of func array becomes 3, so i = 3,. Now when the funcs[j]() is called, it is again using the variable i, which has already been incremented to 3.

Now to solve this, we have many options. Below are two of them.

We can initialize i with let or initialize a new variable index with let and make it equal to i. so when the call is being made, index will be used and its scope will end after initialization. And for calling, index will be initialized again.

Other Option can be to introduce a tempFunc which returns the actual function.

First of all, understand what's wrong with this code:

Here when the funcs[] array is being initialized, i is being incremented, the funcs array is initialized and the size of func array becomes 3, so i = 3,. Now when the funcs[j]() is called, it is again using the variable i, which has already been incremented to 3.

Now to solve this, we have many options. Below are two of them:

We can initialize i with let or initialize a new variable index with let and make it equal to i. So when the call is being made, index will be used and its scope will end after initialization. And for calling, index will be initialized again:

Other Option can be to introduce a tempFunc which returns the actual function:

Source Link

answered Nov 4, 2016 at 7:46

Ali Kahoot

answered Nov 4, 2016 at 7:46

Ali Kahoot

3.6k
2
26
28

First of all, understand whats wrong with this code.

var funcs = [];
for (var i = 0; i < 3; i++) { // let's create 3 functions
 funcs[i] = function() { // and store them in funcs
 console.log("My value: " + i); // each should log its value.
 };
}
for (var j = 0; j < 3; j++) {
 funcs[j](); // and now let's run each one to see
}

Now to solve this, we have many options. Below are two of them.

We can initialize i with let or initialize a new variable index with let and make it equal to i. so when the call is being made, index will be used and its scope will end after initialization. And for calling, index will be initialized again.

var funcs = [];
for (var i = 0; i < 3; i++) { 
 let index = i;
 funcs[i] = function() { 
 console.log("My value: " + index); 
 };
}
for (var j = 0; j < 3; j++) {
 funcs[j](); 
}

Other Option can be to introduce a tempFunc which returns the actual function.

var funcs = [];
function tempFunc(i){
 return function(){
 console.log("My value: " + i);
 };
}
for (var i = 0; i < 3; i++) { 
 funcs[i] = tempFunc(i); 
 
}
for (var j = 0; j < 3; j++) {
 funcs[j](); 
}

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