Question

1. Create a method which accepts two int type variable a and b as parameter and evaluate the following expression: 4.52a+ b and return it. a-b​

Answer 1

Here is a program in C language that accepts two int-type variables 'a' and 'b' as the parameters and evaluates the expression 4.52a + b:

#include <stdio.h>

// Prototype of the Function.

int evaluate_expression(int a, int b);

int main(void) {

 // Declaring two int variables 'a' and 'b'.

 int a, b;

 // Prompt the user to enter two values for the variables 'a' and 'b'

 printf("Enter two values for a and b: ");

 // Reading the values of a and b from the user.

 scanf("%d %d", &a, &b);

 // Evaluating the expression and storing the result in a variable.

 int result = evaluate_expression(a, b);

 // Printing the final result of the expression.

 printf("The result of the expression is: %d\n", result);

 return 0;

}

// Function which evaluates the value of the given expression.

int evaluate_expression(int a, int b) {

 // Evaluate the expression and return the result.

 return (4.52 * a) + b;

}

→ The algorithm to evaluate the value of the expression (4.52a + b):

(1.) Firstly define a function evaluate_expression() that takes two int type variables 'a' and 'b' as input and returns the final result of the expression (4.52a + b).

(2.) In the main() function, declare two int-type variables 'a' and 'b'.

(3.) We would prompt the user to enter two integer values for 'a' and 'b'.

(4.) We would read these values entered by the user using the scanf function.

(5.) We would then pass the values of the variables 'a' and 'b' to the evaluate_expression() function.

(6.) The result of the final expression is then printed using the printf function.

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