Unsigned integers overflow does not "wrap around"

Question

I read following line from Integer Overflow Wiki:

while unsigned integer overflow causes the number to be reduced modulo a power of two, meaning that unsigned integers "wrap around" on overflow.

I have below code where I am trying to create a hash function and got int overflow situation. I tried to mitigate it by using unsigned int but it didn't work and I was able to see negative values.

I know I can handle it other way and it works, as shown in my code comment - Comment 2:. But is it right way and why unsigned int was not wrapping around and overflowing?

int hash(char *word) {
    char *temp = word;
    unsigned int hash = 0; // Comment 1: I tried to handle int overflow using "unsigned" int.
    while (*word != '\0') {
        // Comment 2: This works but I do not want to go this way. 
        //while ((hash * PRIME_MULTIPLIER) < 0) {
        //    hash = (hash * PRIME_MULTIPLIER) + 2147483647;
        //}
        hash = hash * PRIME_MULTIPLIER + *word;
        word++;
    }
    printf("Hash for %s is %d\n", temp, hash);
    return hash;
}

Note that this is a very suboptimal choice for the prime number. You are basically computing `word[0] - word[1] + word[2] - word[3]...` — Dietrich Epp, Feb 16 '16 at 21:57
@ChrisBeck Updated. I meant I was able to see negative values. — hagrawal7777, Feb 16 '16 at 22:00
@hagrawal: Why is your return type `int` if you want an unsigned result? — Blastfurnace, Feb 16 '16 at 22:02
Oh, sorry. I thought your `PRIME_MULTIPLIER` was 2147483647. — Dietrich Epp, Feb 16 '16 at 22:02
Dude, you're going to "see negative numbers" ... if you use a *SIGNED* format specifier like `printf ("%d", i)`! Substitute `'%u"!` Here is a good link on "printf" and format specifiers: http://www.cplusplus.com/reference/cstdio/printf/. ALSO: change your "hash()" return value to UNSIGNED: `unsigned hash(char * word)` — paulsm4, Feb 16 '16 at 22:03
@Blastfurnace That's not the concern. dbush has nailed it with his answer. — hagrawal7777, Feb 16 '16 at 22:04
@hagrawal: Your function will return negative values. Is that what you want? — Blastfurnace, Feb 16 '16 at 22:05
@Blastfurnace Easy. That's what not my question was for. My code may have 100 other things because it could be my testing code. My concern was overflow values when I was printing and dbush nailed the same. — hagrawal7777, Feb 16 '16 at 22:08

score 5 · Accepted Answer · answered Feb 16 '16 at 22:01

5

You're using the wrong format specifier for printf. For an unsigned int, you should be using %u instead of %d.

Also, you should be returning an unsigned int instead of an int.

answered Feb 16 '16 at 22:01

dbush

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Ah, thank you so much buddy. Format specifier was the culprit. I was not worried about return type though. – hagrawal7777 Feb 16 '16 at 22:08
Buddy, one small doubt - at the end of my while loop, `hash` will have a positive value because it is `unsigned int`, so even though when I was using %d format specifier, how come negative value was getting printed. Ideally, unsigned int means there is no bit to hold sign value, then how come negative value was getting printed? – hagrawal7777 Feb 16 '16 at 22:20
1

@hagrawal It was interpreting the unsigned value as a signed value. For example if `hash` contains `0xFFFFFFFF`, using `%d` will output `-1` while `%u` will output `4294967295`. – dbush Feb 16 '16 at 23:05
Please correct me on my understanding - this means when MSB of my unsigned int would be 1 then %d would consider it as negative value? – hagrawal7777 Feb 16 '16 at 23:15

Unsigned integers overflow does not "wrap around"

1 Answers1