Question

Please explain the following so I can do them for myself, thank you very much

If the following C program were to be compiled, linked, and executed, it would print the value of x . How many zeros would appear to the left of the 1 ?

#include <stdio.h> int main()    {     int x = 1234 ;     printf ( "%15.6d\n", x ) ;     return 0 ;     }

The answer is 2

If the following C program were to be compiled, linked, and executed, it would print the value of  x . How many digits would appear to the right of the decimal point?

#include <stdio.h> int main()   {   double x = 123.456789 ;   printf ( "%20.4f\n", x ) ;   return 0 ;   }

Answer is 4

Answer 1

If the following C program were to be compiled, linked, and executed, it would print the value of x . How many zeros would appear to the left of the 1 ?

#include <stdio.h> int main()    {     int x = 1234 ;     printf ( "%15.6d\n", x ) ;     return 0 ;     }

Explanatioon:-

here %15.6d in prntf statement, 15 represent the number of space before the integer value and the .6 represent the what should bbe the total integer length. here x have 4 digits, so it will add 2 zero before the integer so the final value will be 001234

If the following C program were to be compiled, linked, and executed, it would print the value of  x. How many digits would appear to the right of the decimal point?

#include <stdio.h> int main()   {   double x = 123.456789 ;   printf ( "%20.4f\n", x ) ;   return 0 ;   }

Explanation:-

here .4f represent the number of decimal point you want after the decimal places, so here .4 is mentioned so, it will print up to 4 decimal places.