4 ms·

I would assume that is because the hash table takes in a “hash this key” function pointer and a “compare these keys” function pointer, and those must contain th

I would assume that is because the hash table takes in a “hash this key” function pointer and a “compare these keys” function pointer, and those must contain the size for variable-length keys. So even if your user-supplied functions know the length, or only care about the first byte, they have to conform to that API, and then you get an unused parameter.

zabzonk 2y ago
Have functions with different names and different numbers of parameters?
- Sesse__ 2y ago
  And then have a union between the two different kinds of pointers, and a tag to tell which one is in use, then test that tag everywhere? Why? What good does it bring for that extra complexity?
  - zabzonk 2y ago
    Doesn't the name of the function select for the types and use of the parameters? But I haven't written in C for a long time, and this I think could be done more clearly in C++.
    - Sesse__ 2y ago
      You're saying you would write the entire hash table four times, with everything duplicated and no reuse? Again: Why? What do you gain compared to one line to suppress a warning?
      - zabzonk 2y ago
        Clarity? And in C++ at least, I would re-write the functions.
- saagarjha 2y ago
  That would be a bad API design.
- detaro 2y ago
  The point is that you can pass any of the hash functions (or your own compatible) to the datastructure, so it'll obviously call of them with the same set of parameters. And hash functions that need all of them will ignore them.