What does #QNAN0 mean in C code?

Question

C++ / C#

I made a simple program in C. As written in the textbook, I used the conversion specification %f (decimal floating point number) when displaying results where necessary. At the output, I received #QNAN0 where I indicated this specification. What does this mean? And where is my mistake in the code?

#include <stdio.h>

/*Функция расчёта параметров окружности*/

main()
{
    int r, d, l, s, pi;    /*Объявляем пременные*/

    pi = 3.14159;
    printf("Enter radius of your circle:\n");
    scanf("%d", &r);
    printf("%d is the diameter of your circle\n", d = 2 * r);
    printf("%f is the perimeter of your circle\n", l = 2 * pi * r);
    printf("%f is the square of your circle\n", s = pi * (r * r));

    return 0;
}

Command line

Enter radius of your circle:
10
20 is the diameter of your circle
-1.#QNAN0 is the perimeter of your circle
-1.#QNAN0 is the square of your circle

Process returned 0 (0x0)   execution time : 2.289 s
Press any key to continue.

Answer 1

Because printf expects %f ---> double
and you slip it a signed integer int
And the number pi is equal to three.

double d, l, s, pi;    /*Объявляем пременные*/
    int r = 0;
    pi = 3.14159;
    printf("Enter radius of your circle:\n");
    scanf("%d", &r);
    printf("%.2f is the diameter of your circle\n", d = 2 * r); // %.[кол-во знаков]f
    printf("%.2f fis the perimeter of your circle\n", l = 2 * pi * r);
    printf("%.2f is the square of your circle\n", s = pi * (r * r));

Enter radius of your circle: 10
20.00 is the diameter of your circle
62.83 fis the perimeter of your circle
314.16 is the square of your circle