Write a program that uses the conversion specifier g to output the value 9876.12345. Print the value with precisions ranging from 1 to 9.

Written in C++

#include <stdio.h>

int main(){

   double value = 9876.12345;

   printf("1: %.1g\n", value);

   printf("2: %.2g\n", value);

   printf("3: %.3g\n", value);

   printf("4: %.4g\n", value);

   printf("5: %.5g\n", value);

   printf("6: %.6g\n", value);

   printf("7: %.7g\n", value);

   printf("8: %.8g\n", value);

   printf("9: %.9g\n", value);

   return 0;

}

Explanation:

This declares and initializes value as double

   double value = 9876.12345;

This prints the first precision

   printf("1: %.1g\n", value);

This prints the second

   printf("2: %.2g\n", value);

This prints the third

   printf("3: %.3g\n", value);

This prints the fourth

   printf("4: %.4g\n", value);

This prints the fifth

   printf("5: %.5g\n", value);

This prints the sixth

   printf("6: %.6g\n", value);

This prints the seventh

   printf("7: %.7g\n", value);

This prints the eight

   printf("8: %.8g\n", value);

This prints the ninth

   printf("9: %.9g\n", value);

The precision is printed using the following syntax: printf("%.ag\n", value);

Where a represents the number of digits to print

answer; the above statement is (false); a computer (word) is the amount of data;