Freshers' C Class 9 tasks added

Freshers/C/2022_02_20_CClass-9/README.md

@@ -16,9 +16,110 @@ February 19th, 2022
 
 ## Tasks:
 
-*[1st complete [last class'](https://cc-mnnit.github.io/2021-22-Classes/Freshers/C/2022_02_19_CClass-8/) tasks.*
+*[First complete [last class'](https://cc-mnnit.github.io/2021-22-Classes/Freshers/C/2022_02_19_CClass-8/) tasks.*
 
-... coming soon ...
+*Do the following tasks, for more questions you can refer to the book "C in Depth"*.
+
+1. Input two numbers and find their sum using their pointers.
+
+2. As we talked that arrays are a pointer to the first element of the array, write a program to find the sum of the elements of an array using pointers.
+
+3. Write a program to verify that the following ways of accessing 3rd element of an array are the same:
+
+    - array[2]
+    - *(array+2)
+    - 2[array]
+
+4. What is the output of the following code?
+
+    ```cpp
+    #include <stdio.h>
+    int f(int x, int *py, int **ppz) {
+        int y, z;
+        **ppz += 1; 
+        z  = **ppz;
+        *py += 2;
+        y = *py;
+        x += 3;
+        return x + y + z;
+    }
+
+    void main() {
+        int c, *b, **a;
+        c = 4;
+        b = &c;
+        a = &b; 
+        printf( "%d", f(c,b,a));
+    }
+    ```
+
+    <details><summary>Click to see answer...</summary>19
+
+    Hint: it is combination of call by value and call by reference. The value in variable `x` in function `f` is not at same address as that of variable `x` in function `main`.</details>
+
+5. Find the output of the following code snippet:
+
+    ```cpp
+    #include<stdio.h>
+    int main() {
+        int arr[4]={10,20,30,40};   // Assume base address of arr is 5000
+        int x=100,*ptr=arr;
+        printf("%u%d    %d\n", ptr, *ptr, x);
+        x=*ptr++;
+        printf("%u%d    %d\n", ptr, *ptr, x);
+        x=*++ptr;
+        printf ("%u %d  %d\n", ptr, *ptr, x);
+        x=++*ptr;
+        printf ("%u%d   %d\n" ,ptr, *ptr, x);
+        x=(*ptr)++;
+        printf("%u%d    %d\n", ptr, *ptr, x);
+    }
+    ```
+
+    <details><summary>Click to see answer...</summary>
+
+    ```
+    5000 10 100
+    5002 20 10
+    5004 30 30
+    5004 31 31
+    5004 32 31
+    ```
+
+    </details>
+
+6. What is the output of the following C code? Assume that the address of x is 2000 (in decimal) and an integer requires four bytes of memory.
+
+    ```cpp
+    #include <stdio.h>
+    int main()
+    { 
+    unsigned int x[4][3] = {{1, 2, 3}, {4, 5, 6}, 
+                            {7, 8, 9}, {10, 11, 12}};
+    printf("%u, %u, %u", x+3, *(x+3), *(x+2)+3);
+    }
+    ```
+
+    <details><summary>Click to see answer...</summary>
+
+    `2036 2036 2036`
+
+    **Explanation:**
+
+    Given x = 2000
+
+    Since x is considered as a pointer to an array of 3 integers and an integer takes 4 bytes, value of x + 3 = 2000 + 3*3*4 = 2036
+
+    The expression, *(x + 3) also prints same address as x is 2D array.
+
+    The expression *(x + 2) + 3 = 2000 + 2*3*4 + 3*4 = 2036
+    </details>
+
+7. Read about [dynamic memory allocation](https://www.geeksforgeeks.org/dynamic-memory-allocation-in-c-using-malloc-calloc-free-and-realloc/) and [file handling in C](https://www.geeksforgeeks.org/basics-file-handling-c/).
+
+<hr>
+
+This is a reminder to keep practicing questions from CodeForces/CodeChef/HackerRank/HackerEarth.
 
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