// Performs addition, subtraction, multiplication or division depending the input from user
# include <stdio.h>
int main() {
char operator;
double firstNumber,secondNumber;
printf("Enter an operator (+, -, *,): ");
scanf("%c", &operator);
printf("Enter two operands: ");
scanf("%lf %lf",&firstNumber, &secondNumber);
switch(operator)
{
case '+':
printf("%.1lf + %.1lf = %.1lf",firstNumber, secondNumber, firstNumber + secondNumber);
break;
case '-':
printf("%.1lf - %.1lf = %.1lf",firstNumber, secondNumber, firstNumber - secondNumber);
break;
case '*':
printf("%.1lf * %.1lf = %.1lf",firstNumber, secondNumber, firstNumber * secondNumber);
break;
case '/':
printf("%.1lf / %.1lf = %.1lf",firstNumber, secondNumber, firstNumber / secondNumber);
break;
// operator doesn't match any case constant (+, -, *, /)
default:
printf("Error! operator is not correct");
}
return 0;
}Enter an operator (+, -, *,): *
Enter two operands: 1.5
4.5
1.5 * 4.5 = 6.8
The * operator entered by the user is stored in the operator variable. And, the two operands, 1.5 and 4.5 are stored in variables firstNumber and secondNumber respectively.
Since, the operator * matches the case case '*':, the control of the program jumps to
printf("%.1lf * %.1lf = %.1lf",firstNumber, secondNumber, firstNumber * secondNumber);
This statement calculates the product and displays it on the screen.
Finally, the break; statement ends the switch statement.
First read the algorithm, then study the program code line by line. After that, compare the code with the output and finally go through the explanation. This approach helps learners understand both the logic and the implementation properly.
After understanding this example, try to rewrite the same program without looking at the code. Then change some values or logic and run it again. This helps improve confidence and keeps learners engaged on the page for longer.