Binary Numbers. Error When Checking to be sure binary input is binary

Question

I'm writing a simple loop to make sure my input is a valid binary. Ex: I want to throw an error when any number higher than one is user input. I know I need to check against the ASCII numbers. What is going on here? I should not be getting an error when I input binary. Any thoughts?

for (int i=0;i<size;i++)
{
    printf("%i is string sub %i\n",int(binary[i]),i);
    if (int(binary[i]) != 48 || int(binary[i]) != 49)
    {
        printf("ERROR NOT A BINARY NUMBER\n");
        exit(0);
    }
}

input:

Output:

48 is string sub 0
ERROR NOT A BINARY NUMBER

without the definition of binary[] this question cannot be answered. — msw
– msw, Commented Sep 18, 2010 at 3:42

Will A · Accepted Answer · 2010-09-18 03:41:22Z

6

You need to use if (int(binary[i]) != 48 && int(binary[i]) != 49) - note && rather than ||. As it stood, the if(...) was effectively if(true) as binary[i] could not be both 48 and 49 simultaneously.

answered Sep 18, 2010 at 3:41

Will A

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1 Comment

Will A Over a year ago

np - glad to be of assistance.

identity · Accepted Answer · 2010-09-18 03:43:06Z

1

You are using the wrong boolean operator. If you read your if statement out loud, this is what it sounds like:

Execute what is inside the if statement if binary[i] is not '0' OR if it is not '1'.

How can it be two things at the same time?

Use && instead of ||.

answered Sep 18, 2010 at 3:43

identity

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