They're just lovely to see that and check out the violin in case of Cartland.
Now the syntax on vinyl is pretty simple.
So for a while you just need to use vinyl followed by backward open and close.
Now inside this you have to simply define the condition such as is smaller then let's c then our eye
is more than equal to 10.
Now what I see is a counter variable that will help you to continue with the loop till this condition
becomes false.
So we have to simply initialize the value.
So here I simply initialize the Gounder video.
But I see I call on in equal to let's see when inserted the value where they really do.
I mean the same here that I do print all the values.
Let's see if print line.
In the value of i.
Right.
So this is the syntax of why you now are the largest increment of value of i by using the unity operator
of a listless.
And now let's just read the code and see.
So here we go in the output we get one two three four five six seven eight nine and 10.
Now suppose if I ask you to write a program to print all the even numbers starting from and then they'll
then.
So what you do in that case.
So here what they will do I would simply write a if condition.
Let's see if I Margulis two is equal to equal to zero.
Then in that case simply put in a value.
And finally Ingram's in the counter by when I.
So let us know what are called.
So here we get to 4 6 8 and 10.
So this was all about the While loop in case of Cartland.
And now if you are a big In case of programming then please watch this video.
Otherwise if you know how does available works then simply jump on to the next video.
Right.
So as a beginner please watch this video now is an index of light.
So here while the conviction.
And then we put our record here and then increment or decrement our counter variable.
So this is a syntax of value.
Now let us see an example on the left hand side.
So here I have the counter variable A and equal to zero.
Now I have available within which I have some condition I is smarter than three.
Now if I is smaller than three then only execute the code present inside the loop and simply print hello.
Right now at the end there just increment the counter value.
Let's see by when by simply using ipe listless.
Now how does of-I loop.
Now if you underscored then it will simply print hello two times in the output by using loop when loop
2 and loop 3 as usual.
And at the end the loop will terminate.
Right.
So the control flow for the loop is like this.
First check the condition and then execute the call and then implement the counter.
So when the first group is executed the value of i is actually zero.
So zero is smaller then which is actually true service simply print hello in the output console and
at the end it will simply increment the counter value by 1.
So I will become.
Now when I write with the help of this I priceless.
Now this value of equal to when will now appeared at the very initial point of the loop to where the
condition will be checked.
One is smaller than three which is again true.
Again.
Hello in the output console.
And then I place bliss.
And finally it becomes two.
Now this value of i and i just do again comes in the third loop where the condition again becomes true
because two is smaller then three right.
And again we print Hello I'm blessed blessed.
Then again at the end of the loop I becomes three.
Now here comes the twist since it becomes three.
So this will again try to initiate the for loop there.
That condition will actually become false because 3 is not smaller than three right.
So the condition becomes false and the loop dominates and we get only three times hello in the output
console.
So this is how the available works in case of Cartland.