Question about an exercise

Programming
1

I was doing an exercise of the book Programming in C 4th edition, This is the exercise

/*
 * 4. Write a program that acts as a simple “printing” calculator. The program
 * should allow the user to type in expressions of the form
 * 
 *		number   operator
 *
 * The following operators should be recognized by the program:
 *
 *		+     -     *     /     S     E
 *
 * The S operator tells the program to set the “accumulator” to the typed-in
 * number. The E operator tells the program that execution is to end. The
 * arithmetic operations are performed on the contents of the accumulator with
 * the number that was keyed in acting as the second operand. The following is
 * a “sample run” showing how the program should operate:
 *
 *		Begin Calculations
 *		10 S           Set Accumulator to 10
 *		= 10.000000    Contents of Accumulator
 *		2 /            Divide by 2
 *		= 5.000000     Contents of Accumulator
 *		55 -           Subtract 55
 *		-50.000000
 *		100.25 S       Set Accumulator to 100.25
 *		= 100.250000
 *		4 *            Multiply by 4
 *		= 401.000000
 *		0 E            End of program
 *		= 401.000000
 *		End of Calculations.
 *
 * Make certain that the program detects division by zero and also checks for
 * unknown operators.
 *
 * By Faisal Saadatmand
 */

And for some reason when i try to exit typing E, neither my code or the code of a solution that i found on the internet worked, it always end in an infinite loop.
This is the solution i found on the internet:

#include <stdio.h>

int main(void)
{
	float number, accumulator;
	char operator;

	printf("Begin Calculations\n\n");
	scanf("%f %c", &number, &operator);

	accumulator = 0;
	while (operator != 'E') {
		switch (operator) {
		case 'S':
			printf("= %f\n\n", accumulator = number);
			break;
		case '+':
			printf("= %f\n\n", accumulator += number);
			break;
		case '-':
			printf("= %f\n\n", accumulator -= number);
			break;
		case '*':
			printf("= %f\n\n", accumulator *= number);
			break;
		case '/':
			if (number == 0)
				printf("Division by zero\n\n");
			else
				printf("= %f\n\n", accumulator /= number);
			break;
		default:
			printf("Uknown operator\n\n");
			break;
		}
		scanf("%f %c", &number, &operator);
	}
	
	printf("= %f\n\n", accumulator);
	printf("End of Calculations\n\n");

	return 0;
}

This is my solution:

#include <stdio.h>



int main(void){
    float accumulator = 0, number;
    char operator;

    printf("Begin Calculations\n");
    scanf("%f %c", &number, &operator);

    while (operator != 'E'){
        switch (operator){
        case 'S':
            if (accumulator != 0)
                accumulator = number;
            else
                accumulator += number;
                scanf("%f %c", &number, &operator);
            break;
        case '*':
            accumulator *= number;
            printf("= %f\n", accumulator);
            scanf("%f %c", &number, &operator);
            break;
        case '/':
            if (number == 0)
                printf("Cannot divide by 0!");
            else
                accumulator /= number;
                printf("= %f\n", accumulator);
                scanf("%f %c", &number, &operator);
            break;
        case '+':
            accumulator += number;
            printf("= %f\n", accumulator);
            scanf("%f %c", &number, &operator);
            break;
        case '-':
            accumulator -= number;
            printf("= %f\n", accumulator);
            scanf("%f %c", &number, &operator);
            break;
        case 'E':
            break;
        default:
            printf("Unkown operator\n");
            scanf("%f %c", &number, &operator);
            break;
        }
    }
    printf("= %f\n", accumulator);
    printf("End of Calculations\n");

    return 0;
}
2

I’m not sure if this is the right place for questions like this that are not related to Exercism. But I’ll give it a go:

Did you by any chance try to end the calculation by entering a single E without a number?

The function call scanf("%f %c", ...) is very specific about what the program should read: First a floating point value, then some optional whitespace, then a single character.

When you enter E alone the scanf("%f %c", ...) sees the 'E' and cannot process it because it expects a floating point value first. It leaves the 'E' in the input stream and returns (with the return value 0). The number and operator do not get modified, they keep their old values, and the next iteration of the while loop effectively repeats the previous operation.
Then scanf(%f %c", ...) gets executed again, cannot process the 'E' that’s still present in the input stream, returns and leaves number and operator unmodified. The while loop repeats the previous operation.
Then scanf(%f %c", ...) gets executed again, cannot process the 'E' that’s still present in the input stream, returns and leaves number and operator unmodified. The while loop repeats the previous operation.
That’s an infinite loop.

If you want to end the calculation you have to enter a number and then the letter E. You can see that in the example.

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