I want to make simple calculator, that counts problems you entered, when user enters 0_0 it prints results of problems user wrote.

int a[20], b[20], result[20];
int n = 0;
char operation[20]; // 
printf("Enter:\n");



while(1)
{
    if(a[n] == 0 && operation[n] == '_' && b[n] == 0) // Even user types 0_0 cycles does not breaks
    {
        
        break;
    }
    scanf("%i%c%i", &a[n], &operation[n], &b[n]);
    
    n  ;
    
}

Why even if I type 0_0 cycles does not breaks. Whats wrong with a[n], b[n], operation[n] thanks a lot in advance for answer!

CodePudding user response：

You could have spotted the mistake with a simple dry run.

• Initially n = 0 and a[0], b[0], operation[0] have garbage values.
• Now, the control enters the loop, the condition inside if is evaluated. This condition evaluates false because of the above point.
• Now, you're reading the values entered by the user.
• Say, now n = 0, and a[0] = 0, operation[0] = '_' and b[0] = 0 (after entering the values).
• But now, n gets incremented to 1.
• The if condition again evaluates to false because n is not 0 now.

This continues for ever (possibly).

CodePudding user response：

The if statement is placed before you take the users input. So the if never checks the user input but instead checks some uninitialized array values (because n is incremented just after the scan).

You need to read input first and then do the if. So simply swap the if and the scanf

if (scanf("%i%c%i", &a[n], &operation[n], &b[n]) != 3)
{
    // Illegal input. Add error handling here... or simply exit like:
    exit(1);
}

if(a[n] == 0 && operation[n] == '_' && b[n] == 0)
{
    
    break;
}

n  ;

Further notice that you need to make sure that the while stops when n reach 20 (i.e. to avoid writing outside the arrays). Perhaps do

 while(1) --->  while(n < 20)