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Hey, guys, what's up?



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So today we are swallowing one when dusting problem so that proneness treatment is given the most given



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numbers.



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Slap them.



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For example, let's say doing a massage is five and B's seven.



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So by stripping, I mean, is the new video fair is seven and B is five.



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OK, so the venues of A&E are slapped with each other.



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So how against all this problem?



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So what we can use is, first of all, we were doing the most and and B, we will take input from the



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user.



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What other do they lose?



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OK.



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Then what do we do?



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We will take one temporary variable and dump families.



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Atambayev anyway.



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And it will store live in the OWFI.



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So let's see, is five and B is seven.



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So I am what I am doing.



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I am making a temporary variable and this temporary variable will also hold the five.



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I cooperate disvalue here.



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OK, then what I will do, I will right.



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It was be so cooperative and will be in to it.



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So here it will become seven.



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OK, we have to copy the value.



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So the new video fee is seven.



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Then what do we do.



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We will write be it was temp.



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So what is the new value of B new value.



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B is five and Dukan.



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And finally, you can see yourself the new radio phase seven and the new valve BS five.



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So the numbers are slabbed.



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Initially it was five and B was seven.



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OK.



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So with the help of these three lines, I can slap two numbers.



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OK.



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Let's take one more example.



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Suppose the value of is to add the value of Beastie so badly off a stool, then lay off.



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B is T.



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Now what I'm doing.



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I am creating a temporary variable.



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So this is my time.



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Very valuable.



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And this is holding the value of a.



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This is basically copy the value of a temp.



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Copy the value of it.



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And don't temp.



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So too is here.



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Now that we are doing copy the value of B in to it.



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So copy the value of be in to A.



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So then you and your phase three.



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Now what our doing.



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Copy the value of temp in to be complementary of temp.



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And to be so new.



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Any of these two.



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So finally it becomes T and B becomes two.



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So then the Mozart swept.



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OK.



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So let's try it called.



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Let's name it.



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So Abdool numbers.



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NBPA.



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So what do we have to do?



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OK, so first of all, what do we do?



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We will take Goonan marriage and but.



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So let's say I do it on my side, A and B, let us take input from the user.



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So first of all, let us sprint the car into a low fare and we'll see you out before slapping.



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So before slapping the value of A and B are so C out it was.



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And similarly, Soucy out a push, a



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hard line and see out.



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We put.



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So this is the case before swapping.



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Now they will swap for swapping, but they do.



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We will take a very, very temp, which is initially a copy of a new offering to be then.



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But I will do I will copy the value of B into A.



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And then copies of any of them.



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And to be OK.



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Now let us print these values again.



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After swapping.



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So after slapping.



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Now, this is all that we have to do to solve this problem.



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So let us test our program.



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So let's say the value of A and B are five and seven.



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So before setting the radio phase five, after snapping the value of a seven, before stepping the value



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of these life, after setting the value of these SODI, before setting the value will be seven.



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And after us, I think it becomes five.



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OK.



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So the values are getting swept.



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Nowadays, one more way in which we can do a swapping, and that is with the help of Zord, Rick Lazio,



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Peter Zord.



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OK, so what we will do.



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I suppose the value of is five and the value of these seven.



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I will write a quiz is ought to be OK.



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So a quiz.



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The current value of phase five and these seven.



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So this is the new video feed.



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Then I will write be equals is or be.



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So be it was.



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What is the value of the value of phase five or seven.



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Then take Zschau Revd B and R this B seven.



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So seven.



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Now seven a.m. and I will cancel out each other BS.



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Five gained right.



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Equals A zone B so a quiz.



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What is the value of the value of eight five zero seven.



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So five is on seven.



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Then I have to take Zah would be.



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And what is B B's five.



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So five.



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Five and five.



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But against each other.



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The new yellowface.



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Seven.



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So you can see here the value of B is now five.



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So value of B is five and the value of A is now seven.



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So the value of it is now seven.



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So values are getting slammed.



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OK.



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So the values are getting swept.



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So let's come and out.



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So what we can do is a quiz.



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Is order B B GWAS is odd is or B and again eight was A czar B.



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OK,



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let's say the value of and B are five and seven.



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So you can see the values out of getting slammed.



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Initially it was safe.



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Now is seven.



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Initially B was seven.



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After I bring it becomes five.



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So the vendors are getting stabbed.



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Led to the radio phase one and the value of bees, too.



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So, again, the values are getting swept.



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OK.



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Now what C++ does, this trapping operation is very common.



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We have an inverted function to slap two numbers.



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The function is slap a call maybe.



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No.



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Now the values of A and B will be slabbed.



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OK, so this is an N word function slap.



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Let's say a five and B's seven.



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So the values are getting slapped.



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Five, seven.



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OK, one more time.



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Answer the value of a one and these two.



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So they may lose out of getting seven.



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OK, so this.



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So where did the team method first with the help of M variable.



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Second with the help of bitwise operator tzar.



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And third, we have in both function to heping.



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So I will always use this.



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You can use annual Dubeau to or you can also use this material.



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Always use this sarap a comma B.



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OK.



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So this is all for this video.



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Thank you.