Question

Question with explanation . C program

Exercise 7.1 A friend of yours shows you the following method and explains that if number is any two-digit number, the program will output the number backwards. He claims that if number is 17, the method will output 71 Is he right? If not, explain what the program actually does and modify it so that it does the right thing. Submit Answer! *include <stdio.h> #include<stdlib.h> int main(void) { int number = 71; int lastDigit = number 10; int firstDigit = number/10; printf("1", lastDigit + firstDigit); return EXIT_SUCCESS; }

Answer 1

No it's wrong. Because it just prints 8 which addition of two digits. What he did is, just identified two digits and added them while printing. So it will print 8. For clarification see below screenshot.

main.c clang-7 -pthread -lm -o main main.c ./main 1 #include <stdio.h> #include<stdlib.h> 8 2 3 4 int main (void) { 6 7 8 int number 71; int lastDigit = number%10; int firstDigit = number/10; printf("%i", lastDigit + firstDigit); return EXIT_SUCCESS; 9 10 11 }

To rectify it what I did is, I multiplied the lastDigit with 10(because we divided "number" in the line 7 to find the digit in 10's place). So now we will get 10 into lastDigit and adding first digit which 7 will give 17 as the output. For clarification check out the screenshot below.

main.c > clang-7 -pthread -lm -o main main.c ./main 17 1 #include <stdio.h> #include <stdlib.h> 2 3 4 int main( void ) 5 6 7 8 int number = 71; int lastDigit = number % 10; int firstDigit = number / 10; lastDigit-lastDigit*10; printf ("%i\n", lastDigit+firstDigit ); return EXIT_SUCCESS; 9 10 11 12

Modified code to copy:

#include <stdio.h>
#include <stdlib.h>

int main ( void )
{
   int number = 17;
   int lastDigit = number % 10;
   int firstDigit = number / 10;
lastDigit=lastDigit*10;
   printf ( "%i\n", lastDigit+firstDigit );
   return EXIT_SUCCESS;
}

Hope it helps, if you like the answer give it a thumbs up. Thank you.