Answer:
The answer to the given question can be given as:
The value of *iptr is 7. and the value of iptr is dynamic.
Explanation:
In the c++ code, it is defined that x is an integer variable that assigns a value which is 7. Then we define a pointer variable that is *iptr. This variable holds an address of the x variable. When we print the value of the iptr variable. if we use the expression *iptr to print value of the pointer variable by cout that is used in c++ for pint values. so the value of the iptr is 7. If we use the expression iptr sent to cout so we show the address of the variable x. In the pointer, it manages the addresses of dynamically allocated so the address of the variable is changed on execution time.
Answer:
A union (UNION(x,y)) of the sets Sx and Sy represented by x and y, respectively will perform ________4_________ update(s) of the attribute.
Explanation:
The UNION (x, y) disjoint-set data structure unites the dynamic sets that contain x and y, say Sx and Sy, into a new set. It is called the union of the two sets. Before the union operation, the two sets are disjoint. After the union operation, the representative of the resulting set is some member of Sx and Sy or either Sx or Sy. The sets Sx and Sy are then destroyed to remove them from the union collection S. So, four operations are required.
The problem with the swap function is that it loses the value at the first index, as soon as it gets overwritten by the value at the second index. This happens in the first statement. To fix it, you need a helper variable.
First you're going to "park" the index at the first index in that helper variable, then you can safely overwrite it with the value at the second index. Then finally you can write the parked value to the second index:
var swap = function(array, firstIndex, secondIndex) {
let helper = array[firstIndex];
array[firstIndex] = array[secondIndex];
array[secondIndex] = helper;
};
I hope this makes sense to you.
