Thread: negative exponents (simple math prog)

Hey all, i'm using an outdated book, but first of all there is no mention of using pow().

Code:

/* Write a program that evaluates the following expression and displays the results (remember to use exponential format to display the result):
 *
 * (3.31 x 10^-8 x 2.01 x 10^-7) / (7.16 x 10^6 + 2.01 x 10^-8)
*/

#include <stdio.h>

int main(void)
{
    double num;

    num = (10 / 8 / 8 / 8 / 8 / 8 / 8 / 8 / 8);

    printf("%f ", num);
    
    return 0;
}

Currently getting program to output (this is just a start on the calculation and not getting correct answer);

Code:

0.000000

Also, the previous question in book, here is my solution:

Code:

/* 6. Write a program to evaluate the polynomial shown here:
 *
 * 3x^3 - 5x^2 + 6
 * for x = 2.55
*/

#include <stdio.h>

int main(void)
{
    float x = 2.55, num;

    num = (3 * (x*x*x)) - (5 * (x*x)) + 6;

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

    return 0;
}

program output:

Code:

23.231621

Thanks...i understand I am learning how to use the parenthesis correctly. These questions are out of "Programming in C" third Edition by Stephen Kochan (2004).

In C , "10 / 8 / 8" is zero. While "10.0 / 8 / 8" is not zero.

Look up integer division for why.

Tim S.

Presumably the answer they are looking for is more like this. I have no idea what you are thinking trying to divide 10 by 8 eight times!

Code:

#include <stdio.h>
 
int main()
{
    printf("%.2e\n", (3.31e-8 * 2.01e-7) / (7.16e6 + 2.01e-8));
    return 0;
}