string to integer (implement atoi)

Question

Implement atoi to convert a string to an integer.

Requirements for atoi: The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

The following is my code:

int atoi(const char *str) {
    int sign = 1;
    long long i = 0, j = 0;

    while ((*str) != '\0' && isspace(*str)) {
        ++str;
    }

    if (((*str)!='\0') && ((*str) == '+' || (*str) == '-')) {
        if ((*str) == '-') {
            sign = -1;
        }
        ++str;
    }

    if (((*str) != '\0') && (!isdigit(*str))) {
        return 0;
    }

    while (((*str) != '\0') && (isdigit(*str))) {
        i = i * 10 + (*str - '0');
        j = i * sign;
        cout << j << endl;
        if (j > INT_MAX) {
            return INT_MAX;
        }
        if (j < INT_MIN) {
            return INT_MIN;
        }

        ++str;
    }

    return j;
}

Mogria · Accepted Answer · 2013-03-20 19:16:09Z

up vote 2 down vote accepted

if (((*str) != '\0') && (!isdigit(*str))) {
    return 0;
}

You don't need this condition, because of the condition in the while loop afterwards. If that fails the first time j with value 0 is returned anyway.

And a minimal (and maybe unnecessary) optimization:

if (j >= INT_MAX) {
    return INT_MAX;
}
if (j <= INT_MIN) {
    return INT_MIN;
}

Why look for the next character in the string which would only get appended to the number and would increase the value, if the number already has the maximumk/minimum value possible (INT_MAX/INT_MIN).

edited Mar 20 at 19:16

answered Mar 19 at 15:17

Mogria
2337

	Why look for the next character which only makes the number bigger if you already reached INT_MAX/INT_MIN? Can you explain this sentence more detail. I don't understand. – FihopZz Mar 20 at 18:32
	@FihopZz, I edited my answer, I hope its understandable now (English is not my native language) – Mogria Mar 20 at 19:17
	If the number already has the maximum/minimum value, which is checked by `if (j >= INT_MAX)` and `if (j <= INT_MIN)` in my code, there is no necessary to check the next character. It seems that this is what I do. :-). – FihopZz Mar 20 at 22:40
	In your code you return if `j > INT_MAX`, but you don't return if `j` is already equal to `INT_MAX`, so better IMO would be returning if `j >= INT_MAX` like I suggested. – Mogria Mar 21 at 6:48
	Ooooh~, I see. Thanks so much – FihopZz Mar 21 at 12:38

William Morris · Answer 2 · 2013-03-20 17:42:19Z

You seem to have too many checks for '\0'. They don't add anything. Also too many brackets, but better too many than too few.

Also, is long long guaranteed to be bigger than int? I don't think it is. Here's another version that doesn't need a bigger representation than that of int:

int atoi(const char *str)
{
    int n = 0;
    int sign = 1;

    while (isspace(*str)) {
        ++str;
    }
    if (*str == '-') {
        sign = -1;
        ++str;
    } else if (*str == '+') {
        sign = 1;
        ++str;
    }
    while (isdigit(*str)) {
        if (n > INT_MAX/10) { /* EDIT: protect against overflow */
            break;
        }
        n *= 10;
        int ch = *str - '0';

        if (n > INT_MAX - ch) {
            break;
        }
        n += ch;
        ++str;
    }
    if (isdigit(*str)) {
        return sign == 1 ? INT_MAX : INT_MIN;
    }
    return sign * n;
}

EDIT: I added a check to prevent overflow of the n *= 10

EDIT2: improved to avoid using an unsigned int - it was a hack

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string to integer (implement atoi)

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