Question

Read the following code. Is the description of the order of execution (listed below the code) correct? #include <stdio.h> void printPizzaArea(double pizza Diameter) { double pizzaRadius, pizzaArea; const double pival = 3.14; pizzaRadius = pizzaDiameter / 2.0; pizzaArea = pival * pizzaRadius * pizzaRadius; printf("Pizza area is: %lf inches.\n", pizzaArea); } int main(void) { printPizzaArea(12.0); printPizzaArea(16.0); return 0; }

Answer 1

The answer of the given problem is below:

Explanation about the code given.

There is a two integer variable a and b in which a has value 4. we know that when we create variables in programming it reserves some space in the memory which has some address. so variable a and b will have memory adresses. now p1 is a pointer variable which has stored the address of a, it means p1 is pointing to a. now *p1 means the value stored in the memory address pointed by p1. p1 has pointed a and variable a has value 4 so the line b=*p1 means b will also have the value 4.

Now come to the question.

a.) The correct answer of the 1st question is option 1.

which is "a and b are declared to be integer variable".

b.) The correct answer of the 2nd question is " all four options are correct means 1 2 3 4"

c.) only options 1 and 2 are corrcet.

a and b is integer variable and p1 is declared to be pointer variable these statements are true.

d.) correct answers are option1 and option 2.

which is same as c maens a and b is integer variable and p1 is declared to be pointer variable

Here is the screenshot of the executed code in which u can see what is the output when above runs.

screenshot of code--

#include<bits/stdc++.h> using namespace std; int main() { int a=4,b; int *p1; p1=&a; cout<<p1<<endl; b=*p1; cout<<b; }

Ox7ffefb1c9edo 4 Program finished with exit code 0 Press ENTER to exit console. I

here in first line address pointed by p1 is printed,

and second line value of b which is 4.

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