Factorizing && and || in C

Question

I'm not sure what my title should be. But, I was trying to get a nice validation for my code. Below are the only codes allowed to go in my function: c0001, c0002, c0003, c0004, c0005, C0001, C0002, C0003, C0004, C0005. So, other than these 10 codes, it should not accept it and keep promoting the user for a correct code.

if (strlen(meal_choice) == 5 && (meal_choice[0] == 'C' || meal_choice[0] =='c') && meal_choice[1] == '0' && meal_choice[2] == '0' && meal_choice[3] == '0' && (meal_choice[4] == '1' || meal_choice[4] == '2' || meal_choice[4] == '3' || meal_choice[4] == '4' || meal_choice[4] == '5'))

As you can see from my code above, it's way too long. Any idea how I can shorten my code? Thanks!

regular expressions, matching `"^[cC]000[1-5]$"`, see https://stackoverflow.com/questions/1085083/regular-expressions-in-c-examples — Sulthan, Oct 29 '17 at 16:41
`if (tolower(meal_choice[0]) == 'c' && !strcnmp(meal_choice+1, "000", 3) && meal_choice[4] >= '1' && meal_choice [4] <= '5') { ... }` or something like that. Regular expressions more often belong in the problem space than in the solution space, but in this case they can work too, provided you know what you are doing. I would not recommend them for a beginner. — n. m. could be an AI, Oct 29 '17 at 16:46
bool checkCode(char *enteredCode){if tolower(meal_choice[0]) != 'c') return false;.......................... return true}; — Martin James, Oct 29 '17 at 17:06

Sergey Kalinichenko · Answer 1 · 2017-10-29T18:16:15.487

You can check that a character is a digit in a certain range you can compare the character to endpoints of the range with >= and =<, because digits are assigned consecutive codes:

... && meal_choice[4] >= '1' && meal_choice[4] <= '5'

You can also use sscanf's rudimentary regex-like capabilities for validation, like this:

int res = -1;
sscanf(code, "%*[cC]000%*[1-5]%n",  &res);
if (res == 5) {
    ... // The code is valid
}

The above scans and ignores formatted input (note asterisks after percentage signs), and reads the number of processed characters with %n. If all five characters are processed, the input is considered valid.

Demo.

score 1 · Answer 2 · answered Oct 29 '17 at 18:53

In my humble opinion this is the most readable code, which is the most important aspect you should aim for:

const char* valid_codes[] = {
    "c0001", "c0002", "c0003", "c0004", "c0005",
    "C0001", "C0002", "C0003", "C0004", "C0005"
};

const int valid_codes_size = sizeof(valid_codes) / sizeof(valid_codes[0]);

int is_valid_code(const char* code)
{
    for (int i = 0; i < valid_codes_size; ++i)
    {
        if (strcmp(code, valid_codes[i]) == 0)
            return 1;
    }
    return 0;
}

Factorizing && and || in C

2 Answers2