Question

Problems: 1. Write a program to define the following variables, assign them values, and print their values on screen: Integer x = 20, Float y = 40.45 Double z = 91.51e+5 Char w = A • • For the integer variable x, you need to print it in decimal notations, octal notations and hexadecimal notation. For the double variable y, you need to print it in double notations and scientific notation. 2. Write a program with the following three parts: Part 1: Define a variable of type double and assign it the value 400.2354 then convert the value to an integer value. Part 2: Define a variable to hold the ASCII value of the character 'g' then add 8 to it and print the equivalent character on screen. Part 3: Define an integer variable and assign it the value 20, define a double variable and assign it the value 10.62 then add the two variables and print the result on screen. As follow: // Include any required files Int main(void) // Define any required variables // Statements for solving part 1 // Statements for solving part 2 // Statements for solving part 3 return 0; 3.Write a program to read the value of variable x and find the result of variable y by evaluating the following polynomial equation: y = 3 x 3-5x2+6 where x is an integer variable. The program should produce the following on screen in this example, the user enters 2): Enter x: 2 You entered x = 2 y = 10

Answer 1

1.

#include <stdio.h>

#include <stdlib.h>

int main()

{

int x=20;

float y=40.45;

double z=91.51e+5;

char w='A';

printf("%d %o %x \n",x,x,x);

printf("%f \n",y);

printf("%lf %e \n",z,z);

printf("%c \n",w);

return 0;

}

Output:

20 24 14 40.450001 9151000.000000 9.151000e+06 A

2.

#include <stdio.h>

#include <stdlib.h>

int main()

{

//part 1

double d=400.2345;

int i=(int)d; //converting double to int

printf("%d \n",i);

//part 2

int c=(int)'g';

c=c+8;

printf("%c \n",c);

//part 3

int a=20;

double b=10.62;

printf("%lf \n",a+b);

return 0;

}

Output:

400 o 30.620000

3.

#include <stdio.h>

#include <stdlib.h>

int main()

{

int x;

printf("Enter x: ");

scanf("%d",&x);

int y=3*x*x*x-5*x*x+6;

printf("You entered x=%d \n",x);

printf("y=%d \n",y);

return 0;

}

Output:

Enter x: 2 You entered x=2 y=10