Question

Question:Write a C program to input an integer k. Compute 10^k and store the result in a double variable. Display
the result on the standard output using printf.

Implementation
° The program is named lab4c . c. Use the given template lab4c .c (BELOW ) and fill in your code.

° Assume that the input integer k is small enough so that 10^k can be stored in a double variable
named my_double.

° Display on the standard output the result using the printf statement as follows:
printf( "% . 15lf\n", my_double );
--------------------------------------------------------------

Sample Inputs/Outputs

indigo 152 % a.out
Enter an integer>0
1.000000000000000
indigo 153 % a.out
Enter an integer>3
1000.000000000000000
indigo 154 % a.out
Enter an integer>12
1000000000000.000000000000000
indigo 155 % a.out
Enter an integer>15
1000000000000000.000000000000000
indigo 156 % a.out
Enter an integer>-1
0.100000000000000
indigo 157 % a.out
Enter an integer>-10
0.000000000100000
indigo 158 % a.out
Enter an integer>-14
0.000000000000010
indigo 159 % a.out
Enter an integer>-15
0.000000000000001
indigo 160 % a.out
Enter an integer>—16
0.000000000000000

-----------------------------------------------------------------

#include <stdio.h>

void power_of_ten( receive k and my_double from main function )
{

   /* Add your code here */

}  /* end power_of_ten */

main()
{
   int k;
   double my_double;
   
   scanf( "%d", &k );
   
   power_of_ten( pass k and my_double to the function );
   /* Hint: my_double must be passed by reference (i.e., using a pointer). */
   
   /* Do not modify the following line. */
   printf( "%.15lf\n", my_double );
}

Answer 1

Here I've provided the code.

*** CODE STARTS HERE ***

#include <stdio.h>

void power_of_ten( int k, double *mydouble )
{

/* Add your code here */
double result = 1;
  
// compute power k times
while (k--) {
result = result * 10;
}

*mydouble = result;
} /* end power_of_ten */

main()
{
int k;
double my_double;

scanf( "%d", &k );

power_of_ten( k, &my_double);
/* Hint: my_double must be passed by reference (i.e., using a pointer). */

/* Do not modify the following line. */
printf( "%.15lf\n", my_double );
}

*** CODE ENDS HERE ***

for any queries please do comment.