Answer to Question #291524 in C++ for ali

Question

 Write a complete C program consisting of the main function and a void function. The program uses the void function to display the following menu: 1. Square root of sum 2. Logarithm to base e of sum of squares Please enter your choice The program then prompts for and reads two values x and y each of type double. It then uses a switch statement to do the following, based on the selected choice: • Choice 1: The program displays the value of sqrt(x + y) if x + y is greater than or equal to zero; otherwise it displays an appropriate error message. • Choice 2: The program displays the value of log(x * x + y * y) if x2 + y2 is greater than zero; otherwise it displays an appropriate error message.

Answer 1

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <Windows.h>
#include <math.h>


void print_menu()
{
    printf("1. Square root of sum \n2. Logarithm to base e of sum of squares \nPlease enter your choice : ");
}




int main()
{
    int res = 0;
    print_menu();
    while (1)
    {
        res = 0;
        scanf("%i", &res);
        if (res == 1 || res == 2)
        {
            break;
        }
        printf("Please enter your choice (1 or 2) :");
    }
    
    double x=0, y=0;
    printf("Please enter x : ");
    scanf("%lf", &x);
    printf("Please enter y : ");
    scanf("%lf", &y);


    if (res == 1)
    {
        if (x + y >= 0)
        {
            double tmp = sqrt(x + y);
            printf("Result : %lf\n", tmp);
        }
        else
        {
            printf("X + Y is less than 0\n");
        }
    }
    else if (res == 2)
    {
        if ((pow(x, 2) + pow(y, 2)) > 0)
        {
            double tmp = log(x * x + y * y);
            printf("Result : %lf\n", tmp);
        }
        else
        {
            printf("X^2 + Y^2 is less or equal than 0\n");
        }
    }


    system("Pause");
    return 0;


}