 Hello friends. I'm Sanjay Gupta. In this video, I'm going to demonstrate you how you can count positive, negative and zero elements available in array using pointer. Before starting, you can note my information. You can follow or subscribe my YouTube channel through the URL youtube.com slash Sanjay Gupta underscore tech school. You can download my programming app Techimus, which is available on Google Play. Now I'm going to implement the solution of this problem with the help of C programming. So first I'm going to include the header file STDI dot H. Then inside main function, I'm going to define the code which belongs to the problem. So first time declaring an array, a pointer, then three counter variables C1, C2 and C3. And all three variables are initialized with zero and a variable I. Now with the help of printf, I'm going to display the message enter 10 elements on console to read those elements. I am applying a for loop which will repeat 10 times and it will repeat scan of statement, which will read all 10 elements. And those elements will be stored inside a array. Now I have to process this array with the help of pointer. So for that purpose, I am writing the statement p equals to a. So here name of array is providing base address of array. So you can note that name of array provides base address of the array. So if I write p equals to a, it means base address will be automatically stored inside the p pointer. Now I can access all the locations of array with the help of this p pointer. So for that purpose, I am again applying a for loop inside this for loop. I have to apply the condition that is as to this p greater than zero. So here as to risk p means value at address and the address which belongs to the array. So value of that address will be compared with zero. If it is greater than zero, then I can apply C1 plus plus. Else if as to risk p is less than zero, then I can apply C2 plus plus. Otherwise I can apply C3 plus plus. So this way I have checked elements of array with the help of pointer, whether they are positive, negative or zero. After comparison of first location, I have to move to the next location of array. So for that purpose, I have to apply p plus plus. So p plus plus will increase the base address and the pointer will be pointing to the next location. And then next location will be checked for positive, negative and zero. So this process will repeat 10 times. So 10 different locations will be accessed through as to risk p with the help of p plus plus notation. After completion of this loop, I can print the result on console. So positive equals to Poisson d negative equals to Poisson d and zero equals to Poisson d. Then C1, C2 and C3. So this way I have printed all the calculated result on console that how many positive, how many negative and how many zero elements are available inside array. And you can see in the process part, I have used pointer instead of array name. Now I am going to execute this code so that you can understand how it is working. So I have to enter the 10 numbers. So I am entering mixed information which contains negative, positive and zero elements. So you can see the output positive elements are four negative elements are three and zero elements are also three. So now count positive elements are one, two, eight and six. So these are four in quantity negative are minus nine, minus seven, minus five. These are three and zero is available three times. So the program is working properly. It is counting correct numbers of positive, negative and zero elements available in array. And all the operations of a process is done with the help of pointer. So this way I hope you have understood how we can process array through pointers and how we can count positive, negative and zero elements of array using pointer. If you want to watch more programming related videos, you can follow or subscribe my YouTube channel through the URL, youtube.com slash sanjaygupta underscore tech school. You can download my programming app TechMS which is available on Google Play. Thank you for watching this video.