The condition in the loop specifies i<3. Shouldn't the loop stop at i = 2? if so, shouldn't outside loop be 2 not 3?
Thanks.
for (var i = 0; i < 3; i++){
console.log(i, " loop")
if(i%2===0){
console.log (i,'even numbers in loop ');
}
}
console.log(i, " outside loop")
That's the correct behaviour of loops.
The loop starts with i = 0, then i = 1, then i = 2, then i = 3. oops i is no more less than 3, break the loop. that's what's going on, so when you log the value of i you get 3 and not 2.
Loop: it breaks as soon as i < 3 is not verified, which means in the last iteration. More informations about iterations.
Scope: because i is declared using var, which attaches the variable to the global scope. Using var is highly discouraged, use let instead, which declares variables in block scope. More informations about var and let.
Related
I am having trouble understanding this nested loop and how it works.
function foo() {
function bar(a) {
i = 2; // change value of in in enclosing scope
console.log(a + i);
}
for (var i=0; i<10; i++) {
bar(i); // infinite loop
}
}
foo();
Can someone explain why is bar called with 3? I thought first time bar is called it would be zero from inital value of i in the enclosing for loop?
I know this is an infinite loop where a + i always prints 5
I know that the value of parameter a passed to bar is always 3
I know that the value of i in function bar is always 2
Can someone walk me through one repeitition of this infinite loop, explaining the order in which things occur.
Now the reason why it steps in an infinite loop is because of 'var', i = 2 gets hoisted at the top and whenever the for loop iterates i is reassigned the value 2, therefore a never ending loop.
It is a infinite loop because i is declare with var so it remain at the top of foo function and then bar function modify the same i. So i will be values 0, 2, 3, 2, 3...
function foo() {
function bar(a) {
i = 2; // change value of in in enclosing scope
//console.log(a + i);
}
for (var i = 0; i < 10; i++) {
console.log(i);// always 0, 3, 3....
bar(i); // infinite loop
//console.log(i);// always 2
}
}
foo();
I was trying to make a for loop that increments through the numbers 1 - 4 and print them, but when I printed the value of i after the loop, my code outputs 5.
for (i = 1; i < 5; i++) {
document.write(i + "<br>"); //Outputs numbers 1 - 4
}
document.write("New i: " + i); //Outputs 5
How is this possible if i can only increment up until its value is 4?
When you declare a variable in a loop statement, it stores the variable in the same scope as the loop. Loops increment (i++) at the end of the loop, then check the condition (i < 5) to see if they should repeat. After the loop, the variable i still exists. See the snippet below for a play-by play.
Also, you should use the var keyword when declaring i, otherwise, the variable is stored in the global scope (which is bad practice).
//variable i is captured here (with the var keyword)
for(var i = 1; i < 5; i++) {
//for the fourth iteration, i === 4
//so it prints 4.
document.write(i);
//i++ happens right now, so i now is 5.
//then the loop checks the condition to see if it
//should continue. i isn't less than 5, so the loop breaks.
}
//and the value of i is still 5. so this prints 5.
document.write('</br>' + i);
You could move the final-expression part into the condition part of the for statement, where the increment takes place only if the value is smaller than 4.
var i;
for (i = 0; i < 4 && ++i;) {
console.log(i); // 1 ... 4
}
console.log('last value', i); // 4
The test clause of i < 5 is evaluated at the beginning of each loop; if it is false, the loop exits (leaving the value at 5). The i++ part always happens at the end of an iteration, meaning that after the number 4 has been processed in the loop, it increments it to 5, then it loops to the top of the loop where it checks the condition again (i<5) which it fails, but the value of i is not reversed back to 4.
The condition of the loop block to be processed is i < 5, thus i must be less than 5 for the statements and expressions within it to be evaluated
You increment i by 1 with each pass
When i equals 4, you increment it by 1 again such that i will be equal to 5. i < 5 is no longer true, so within the loop block is not executed
Nothing turns i back one increment; so it retains it after the loop
You can apply unique ways, but generally the first and last example below is the common approach:
var log = console.log;
// Example 1: Decrease Loop Iterator After
for (var i = 1; i < 5; i++)
log(i);
log('New i:', --i); // 4
// Example 2: Count in conditional
var count = 0;
for (var i = 1; i < 5 && ++count; i++)
log(i);
log('New i:', count); // 4
// Example 3: Count in loop body
var count = 0;
for (var i = 1; i < 5 && ++count; i++)
log(i);
log('New i:', count); // 4
you have declared i as global variable.
That's how the for loop works.
for (initialization, condition, increment/decrement ) {
}
You have initialized with 1 and it satisfies the condition i<5 so it enters the loop and prints i. after then it performs the increment/decrement operation and increases the value by 1 in your case. So while i=4 and i<5 it enters the loop and prints 4, but after then it performs increment operation and increase the value of i to 5.
In the next step, as the condition i<5 is not satisfied , so it breaks the loop and go to the last statement, where you are printing i, that is 5.
Hope it is clear now.
var array = [1, 2, 3, 4];
array.forEach(function(value) {
document.write(value + "<br />");
});
document.write("Values processed: " + array.length);
No, I‘m not kidding. Use arrays whenever you can, they will spare you a lot of trouble. The side effect you‘re in fact asking for can cause hard-to-track bugs (i looks like being meant for the loop only, imagine tons of lines between the loop and the last document.write). And there‘s the off-by-one error problem with for loops.
You‘ll probably anyway end up with an array once your code starts doing meaningful stuff (you don‘t want to count up to 4, do you?). So do yourself a favor an embrace all those powerful Array methods.
please use count variable because i value increases after each statement execution. i.e after your loop is finished execution i will be always 5
var count=0;
for (i = 1; i <5; i++) {
document.write(i + "<br>"); //I increases after these statement execution so it will be 5 at last
count=count+1;
}
document.write("New i: " + count); //Outputs 5
Are all of the block scope variables in a for loop block hoisted above the loop header itself?
var x = 4;
for(let i = 3; i < x; i++) {
let x = 2;
...
}
Should this produce a dead-zone error on x every time i is compared to x in the loop header? I understand i is pushed down in the block scope, but why not x?
The relevant section of the language specification is 13.7.4
If the for statement contains a let or const declaration then a scope is created.
Each iteration creates a scope, if the for statement contains a let declaration.
If the for body is a block then a scope is created.
Here are some examples and the scopes that are created:
//no scope
for(i = 0; i<3; i++) console.log(i);
//no scope
for(var i = 0; i<3; i++) console.log(i);
//for scope and iteration scope
for(let i = 0; i<3; i++) console.log(i);
// for scope, iteration scope and block scope
for(let i = 0; i<3; i++) {
console.log(i);
}
Why do we need iteration scopes? For closures:
for(let i = 0; i<3; i++) {
setTimeout(() => console.log(i), 10);
}
Output without iteration scope: 3,3,3. With iteration scope: 0,1,2
based on that, i think let x are only available in that for scope.
Are all of the block scope variables in a for loop block hoisted above the loop header itself?
No, they are hoisted in the block scope
Should this produce a dead-zone error on x every time i is compared to x in the loop header? I understand i is pushed down in the block scope, but why not x?
No, because x in the loop header is referencing x defined as 4 var x = 4
You cannot access a let variable before it has been declared as that time between scoping and declaration is the TDZ.
ex.
console.log(x) // Reference Error, x is in TDZ
let x = 4
console.log(x) // --> 4
I misunderstood, and thought that there would be only one block scope. I balked at taking the two arbitrary variables in the expression i < x and having to decide which block scope to look them up, which is impossible to determine.
The correct approach is to build two nested block scopes, one for the loop header and one for the loop body. That means let i is seen in both the loop header and the loop body scopes, and the let x in the loop is only seen in the loop body scope. Easy peasy, lemon squeezy.
var functions = [];
(function foo() {
for (var i=0; i<5; i++ ) {
var f = function() {
console.log(i);
}
functions.push(f);
}
for (var i=0; i<5; i++ ) {
f = new Function("console.log(i);");
functions.push(f);
}
})();
for (var i=0; i<10; i++ ) {
functions[i]();
}
the output is :
5
5
5
5
5
5
6
7
8
9
can anybody help me sort out the flow?
first of all you should know whenever you want to run this function:
Function("console.log(i);");
this would use the current context's i variable as its value. so you might want to do:
var functions = [];
(function foo() {
for (var i = 0; i < 5; i++) {
var f = function func() {
console.log(func._i);
};
f._i = i;
functions.push(f);
}
for (var i = 0; i < 5; i++) {
f = new Function("console.log(" + i + ");");
functions.push(f);
}
})();
for (var i = 0; i < 10; i++) {
functions[i]();
}
I changed it this way and my result is:
0, 1, 2, 3, 4, 5, 6, 7 ,8, 9
Why is that?
1- all of the functions in your first for loop use a single variable i, and as its last value is 5 then all would print 5.
2- the other ones which get created in your second for loop, whenever they get invoked they would use the i variable in their current context, and as far as they get invoked in your last for loop, they use the i variable in the loop, to make it clear. you can easily change your last for loop like this:
for (var j = 0; j < 10; j++) {
functions[j]();
}
and as there is no i variable in the context (I changed it to j), then your output would be:
5 5 5 5 5 undefined undefined undefined undefined undefined
undefined is there because function can not fin any i variable in the context.
It's actually printing out five 5's and then 5, 6, 7, 8, 9. The first five 5's are from var i in that for loop, the functions save a reference to that var i and access it when you run it, and since it equals 5 at the end of the for loop, they all return 5. The second set of numbers comes from the quoted function accessing i from the for loop there, ergo it printing out 5, 6, 7, 8, 9.
Lots of good answers here:
How do JavaScript closures work?
The first 'for' loop is generating 5 functions each of them creates a closure which captures the 'i' as 5 (last value of 5 for that loop after last increment) so they all render out 5 (it just shows in Chrome as the first 5 with 5 in a circle next to it meaning the same value was logged out 5 times)
The second 'for' loop generates functions that will use 'i' from the scope it executes from, in this case the scope is the 3rd 'for' loop...
Closure allow you to access the variable from parent scope, which sometime cause strange or wrong behavior, here in this scenario closure is causing the output: 5 5 5 5 5 0 1 2 3 4
I am trying this code with:
f = new Function("alert("+i+");");
In first loop:
for (var i = 0; i < 5; i++) {
var f = function func() {
console.log(func._i);
};
f._i = i;
functions.push(f);
}
you are referencing i of parent means loop variable, so five different functions are created all with the same copy of variable i, which is changing with iteration, at the end of loop where condition break is i=5 since all functions are referencing the same copy, so change to original copy will effect all. So each function now has the i=5
Now in second loop, you are appending it as a string, which doesn't refer to original copy but creating a new string for each iteration:
for (var i = 0; i < 5; i++) {
f = new Function("console.log("+i+");");
functions.push(f);
}
this create five different functions with each different string:
console.log("0");
console.log("1");
console.log("2");
console.log("3");
console.log("4");
Now calling each function created by loop1 functions[0] to functions[4] is referring the same copy of i will print 5 while second functions[5]to end has there own strings to print.
I have the following small code snippet with the following expected and real output. My question is quiet simple. Why is it printing it this sequence? and how to I print the expected output?
Gr,
expected result:
0
1
2
0
1
2
real result:
0
1
2
3
3
3
this is the code:
var functions = [];
for (var i=0; i<10; i++) {
console.log (i);
functions.push (function () {
console.log (i);
});
};
for (var j=0; j<functions.length; j++) {
functions[j] ();
};
The functions that you push into the array doesn't log the value of i as it was when the function was created, they log the value of i at the time that the function is called.
Once the first loop ends, the value of i is 10, therefore any of the functions called after that will log the value 10.
If you want to preserve the value of i at different states, you can use a closure to make a copy of the value:
for (var i=0; i<10; i++) {
console.log (i);
(function(){
var copy = i;
functions.push (function () {
console.log (copy);
});
})();
};
The local variable copy will get the value of i and retain the value. You can also pass the value as a parameter to the function:
for (var i=0; i<10; i++) {
console.log (i);
(function(copy){
functions.push (function () {
console.log (copy);
});
})(i);
};
The expected result should be:
1
2
...
10
10
10
... 7 more times
The reason for this is simple. The console.log(i) inside your loop is correctly printing the value of i at each iteration of the loop. When you create and push a function into the functions array, what you're doing is closing each of those functions over the same variable i. At the end of your loop, i no longer satisfies your loop condition, so i = 10 is true. As a result, since each of those functions is going to execute console.log(i), and they're each closed over the same i, which now has value 10, you should expect to see the value 10 printed 10 times.
To prevent this, you will want to make a function which returns a function rather than creating functions directly in a loop:
var functions = [], i, j;
function createEmitter(i) {
return function () {
console.log(i);
};
}
for (i = 0; i < 10; i++) {
console.log(i);
functions.push(createEmitter(i));
};
for (j = 0; j < functions.length; j++) {
functions[j]();
};
Now, each of those created functions is closed over its own private scope variable, which resolves the problem.
You should update your code example to be i < 3 so that your results and function match up.
The functions you push into the functions array are storing a reference to the variable i, which after executing the top loop, is 10. So when it executes, it will go get the variable i (which is 10) and print that 10 times.
Here's a good way to see this in action:
for (var i=0; i<10; i++) {
console.log (i);
};
console.log(i) //=> 10
When you are using a variable, remember that the variable can change, it's not frozen at it's current value. You are only holding on to a reference to something else.
To fix this problem, I would run this type of minor refactor on the code (because the other answers have already created an extra scope, figured I'd give you something different). Rather than storing 10 functions, just store the numbers and execute them with a single function. This is a more elegant way to write it anyway, and takes up less space. I'm sure this example was abstracted from whatever code was really giving you the problem, but the general pattern still applies.
numbers = [];
for (var i=0; i<10; i++) {
console.log (i);
numbers.push(i);
};
numbers.forEach(function(i){
console.log(i);
});