C give memory snapshots of all variables after the following...

C++ Give memory snapshots of all variables after the following statements are executed. For pointer variables, show with arrows what variable they point to. int x = 5, y = 7, z; int *ptr_1 = &y, *ptr_2 = &z, *ptr_3 = &z; *ptr_2 = x; x = *ptr_1 + *ptr_2 + *ptr_3; Give memory snapshots of all variables after the following statements are executed (don't forget the variable k). For pointer variables, show with arrows what variable they point to. int k; double t[6] = {1.0, 2.0, 5.0, 4.0, 7.0, 6.0}, p[6]; double *ptr = &t[0]; for (k = 0; k <= 5; k++) p[k] = *(ptr + k) + k;

Transcript

00:01 In this question, we have to give memory snapshots of the variables that are given to us.

00:14 Like this.

00:16 So, we have to give memory snapshots of all variables after following the statements that are executed.

00:23 So, the pointer variables and the variables that follow the statements that are executed.

00:30 All those are given to us in the question.

00:33 So, we will just see in a basic compiler how we are going to proceed with this as everything is given to us in the question.

00:42 So, we just have to follow that.

00:45 So, first of all, we will, first state of statements that are there will be like this.

00:53 So, here what is done is we have declared three integer variables that is x, y and z.

01:04 So, as you can see, we have initialized x with the value of 5 and y with the value of 7 and z is left uninitialized.

01:18 So, this second line that you can see, there are three integer points that is ptr1, ptr2 and ptr3 are declared over here.

01:27 So, ptr1 is initialized with the address of y and ptr2 is initialized with the address of z and 3 with the address of both of these are initialized with the address of z as you can.

01:42 So, ptr2 is equal to x this assigns the value of x that is 5 to the variable z.

01:49 So, indirectly through the pointer of 2, now z will become 5.

01:55 So, the next set of values that we set is x is equal to ptr1 plus 2 plus 3.

02:03 Here what we have done is the value of x is updated by adding the values pointed to by ptr1 2, 3.

02:13 So, x will become 7 plus 5 that is and also another 5...

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