How does a for loop behave in JS when we call the varriable outside the curly braces?

Question

Please explain the logic, I am confuse, it behaves strangely :(

for(x=0;x<6;x+=2){}
Console.log(x)
//6
_____________________
for(x=1;x<6;x+=2){}
Console.log(x)
//7
______________________
for(x=2;x<6;x+=2){}
Console.log(x)
//6

Answer 1

Well, as a simple explanation :

For the first case :
for (var x = 0; x < 6; x +=2){}
console.log(x);

// output : 6

So, at the very first time x is 0
Then at the second attempt x is increased by 2 ! That means 0 + 2 = 2!
3rd attempt 2 + 2 = 4 !
4th attempt => 4 + 2 = 6

Now x has been 6 and our condition is untill x < 6 we will increase it by 2! Now 6 is not less than 6! So the loop will stop! And that's how the output is 6!

For the second case :
for (var x = 1; x < 6; x +=2){}
console.log(x);

// output : 7

1st attempt => 1 + 2 = 3
2nd attempt => 3 + 2 = 5
The loop will run one more time! Cause 5 is less than 6! So,
Last attempt => 5 + 2 = 7

And that's why the output is 7!

And the 3rd case is as same as above 😇

You will get an error if you use 'let' instead of using 'var' ! Try doing that 😁

Answer 2

ThanksitsQuasar✅ &Diego
It was so simple but.......
Thanks and love you both😍😊

Answer 3

ThanksArb Rahim Badsa
&
Thanks(‎Basel Al_hajeri)('MBH')

Answer 4

Of course, in the middle is increments before the condition

Answer 5

Tahir Usman after every loops the complier runs other one just .
1.the last one in loop is 4 after loop 6
2.the last one in loop is 5 after loop 7
3.the ....in loop is 4 after it 6
As usual 🙂

Answer 6

As Arb Rahim Badsa said
I've made it a simple to make u understanding

Answer 7

for(x=0;x<6;x+=2){}
Console.log(x);

means:
1. 2+2=4
2. 4+2=6
3. 6=6 => nothing

Answer 8

for(x=1;x<6;x+=2){}
Values "x" takes are 1, 3, 5 , 7 (not less than 6 so loop stops).

for(x=0;x=6;x+=2){}
Values "x" takes are 0, 2, 4, 6 (not less than 6 so loop stops).

Answer 9

see the code here

https://code.sololearn.com/Wx7X5qNhdx2x/#js