Error with this code? - c

Question

So, I don't see what is flawed with my logic on this problem.
It reads the expression from left to right, and operands are floating point numbers.
However, my program gets stuck in the loop. It reads and assigns the final number, for example it assigns 10.5 to num1, but it never exits the loop.
int main(void)
{
float num1, num2;
char oper = 0;
printf("Enter an expression: ");
scanf("%f", &num1);
while (oper != ('\n' || EOF))
{
oper = getchar();
scanf("%f", &num2);
switch (oper)
{
case '+':
num1 += num2;
break;
case '-':
num1 -= num2;
break;
case '*':
num1 *= num2;
break;
case '/':
num1 /= num2;
break;
}
}
printf("Value of Expression: %.2f", num1);
return 0;
}
Expected output:
Enter an expression: 1+2.5*3
Value of expression: 10.5

Answer 1

while (oper != ('\n' || EOF)) Here, ('\n' || EOF) == 1, so the loop will not terminate unless you input the ASCII character that corresponds to 1 (which is unprintable). As Leffler points out, this should be while (oper != '\n' && oper != EOF) (though see the next point).
You check oper in your while loop before you read from it, so your while loop will terminate one step "after" the newline/EOF. The while loop should actually be:
while(1) {
oper = getchar();
if(oper == '\n' || oper == EOF) {
break; // Exit out of loop, ignoring the rest of the code inside the loop.
}
scanf("%f", &num2);
// ...
}
getchar returns an int, but oper is a char. Change oper to an int (EOF cannot be represented as a char)

Answer 2

Make oper an int. 255 is not equal to EOF

Answer 3

You need to revise the loop ending conditions. Your program is actually busy waiting for you to type another number after the first newline.
#include <stdio.h>
int main(void)
{
float num1, num2;
printf("Enter an expression: ");
if (scanf("%f", &num1) != 1)
return 1;
printf("Num1: %f\n", num1);
while (1)
{
int oper;
if ((oper = getchar()) == EOF || oper == '\n')
break;
printf("Operator: %c\n", oper);
if (scanf("%f", &num2) != 1)
break;
printf("Num2: %f\n", num2);
switch (oper)
{
case '+':
num1 += num2;
break;
case '-':
num1 -= num2;
break;
case '*':
num1 *= num2;
break;
case '/':
num1 /= num2;
break;
}
printf("num1 = %f\n", num1);
}
printf("Value of Expression: %.2f\n", num1);
return 0;
}
Note that this code checks each input operation to make sure it was successful. It also uses int oper instead of char oper to ensure that EOF is detected accurately. And the key trick is not to ask for more input after the getchar() reads the newline; your program was stuck waiting for you to type something in response to the second scanf() call. Notice how this code prints its inputs so you can tell what's going on. When a program is misbehaving, it is one of the simpler techniques to find out what's going wrong. When I saw Operator: and a blank line — and the program hanging — I knew very quickly what the trouble was.
Example run (program name: stuck):
$ ./stuck
Enter an expression: 1+2.5*3
Num1: 1.000000
Operator: +
Num2: 2.500000
num1 = 3.500000
Operator: *
Num2: 3.000000
num1 = 10.500000
Value of Expression: 10.50
$