Question

6. Consider a C program that reads two real numbers from the keyboard followed by a character where the character can be one of the operators +,-, *, 1, or % providing the addition, subtraction, multiplication, division, or remainder respectively. Then the result is displayed on the screen For example, if the user types 2.0 3.0 % then your code will display: Note: In the above example the inputs are real numbers but the remainder only performs an integer operation. Hence, the two real numbers must be type cast to allow computation by remainder. In the case where an invalid operator is entered (e.g. S) then an error message is printed to the standard output, e.g. Invalid entry! Write the above program by ONLY using if, else if, and/or else statements.

Answer 1

Hello, here is the completed code you wanted. Every important statement is explained using comments. Please check and let me know if you have any doubts. Thanks.

CODE

#include<stdio.h>

int main()
{
   float operand1, operand2;
   char operator;
  
   //prompt for operands and operator
   printf("Enter the operands and operator: ");
   scanf("%f %f %c", &operand1, &operand2, &operator);
  
   //if operator is +
   if(operator == '+')
       //prints output
       printf("%.1f %c %.1f = %.1f", operand1, operator , operand2, (operand1 + operand2));
   //if operator is -
   else if(operator == '-')
       //prints output
       printf("%.1f %c %.1f = %.1f", operand1, operator , operand2, (operand1 - operand2));
  
   //if operator is *
   else if(operator == '*')
       //prints output
       printf("%.1f %c %.1f = %.1f", operand1, operator , operand2, (operand1 * operand2));
   //if operator is /
   else if(operator == '/')
       //prints output
       printf("%.1f %c %.1f = %.1f", operand1, operator , operand2, (operand1 / operand2));
   //if operator is %
   else if(operator == '%')
       //prints output
       printf("%d %c %d = %d", (int)operand1, operator , (int)operand2, ((int)operand1 % (int)operand2));
   //if wrong operator
   else
       printf("Invalid entry!");
   return 0;
}

OUTPUT

nter the operands and operator: 2.0 3.0 Process exited after Press any key to continue -- - 5.622 seconds with retu rn value 0

nter the operands and operato: 2.0 3.0 + 2.0 3.0 5.0 Process exited after 7.799 seconds with return value 0 Press any key to continue - -

Enter the operands and operator: 2.0 3.0 2.03.0 .0-3. =-1.0 rocess exited after 4.354 seconds with return value D Press any key to continue - -

nter the operands and operator: 2.0 30* 2.0 3.0-6.0 rocess exited after 5.546 seconds with return value 0 Press any key to continue

Enter the operands and operator: 2.0 3.0/ 2.0 3.0 = 0.7 Process exited after 5.345 seconds with return value 0 Press any key to continue - - -

nter the operands and operator 2.0 3.0 Invalid entry! rocess exited after 7.645 seconds with return value 0 ress any key to continue . -

CODE SCREEN SHOT

1 include<stdio.h> 2 int main() loat operandi, operand2 char operator //prompt for operands and operator printf(" Enter the operands and operator: ") scanf ("%f %f %c", &operandi , &operand2 , &operator); 10 12 //if operator is t if (operator+ 13 14 //prints output 15 printf ("8.1f 동с 동.1f = 똥.1f", operandi , operand2 , (operand1 + operand2)); operator , 16 //if operator is else if (operator-) //prints output 17 18 printf ( "8.1f 동с 동.1f = 똥.1f", operandi , operand2 , (operandi - operand2)); operator , 20 21 //if operator is * else if (operator*) //prints output 23 24 printf ( "8.1f 동с 동.1f = 똥.1f", operandi , operand2 , (operand1 * operand2)); operator , 25 //if operator is / else if (operator) //prints output 26 27 28 printf("동.1f 동с 동.1f = 똥.1f", operandi , operand2 , (operand1 / operand2)); operator , 29 //if operator is 울 30 else if (operator- ') 31 //prints output printf("Sd%cSd //if wrong operator 32 %d" , (int) operandi, (int) operand2, ((int) operand1 % (int) operand2)); operator = , 34 else printf ("Invalid entry!") 35 36 return 0; 37 L