Hash pseudo code does not match actual code

[byarmis]

The pseduo-code

    hash = 0
    string_len = length(string)
    for i = 0, 1, ..., string_len:
        hash += (a ** (string_len - (i+1))) * char_code(string[i])
    hash = hash % num_buckets
    return hash

takes the modulus of the hash value and the number of buckets once, immediately before returning it. The actual code

static int ht_hash(const char* s, const int a, const int m) {
    long hash = 0;
    const int len_s = strlen(s);
    for (int i = 0; i < len_s; i++) {
        hash += (long)pow(a, len_s - (i+1)) * s[i];
        hash = hash % m;
    }
    return (int)hash;
}

takes the modulus of the hash value and the number of buckets for each letter in the input string.

[SimonWang2014]

yep, the pseudocode is a little difference with the actual code, but the results are equally :)

[cyber-chris]

The results are not equal. Run this and see it outputs 21, not 5 as in the pseudo-code. Although, it's probably because of C precision vs Python precision.

#include <stdio.h>

// hash_table.c
static int ht_hash(const char* s, const int a, const int m) {
    long hash = 0;
    const int len_s = strlen(s);
    for (int i = 0; i < len_s; i++) {
        hash += (long)pow(a, len_s - (i+1)) * s[i];
        hash = hash % m;
    }
    return (int)hash;
}

int main()
{
    printf("Hello World, %d", ht_hash("cat", 163, 53));

    return 0;
}