Radical Language Modification

[Duthomhas]

Hey all, I'm playing around with bignums again...

I just realized I could do something cool and make code that looks like this:

d = a + b (mod c);

...where only b (and, of course, d) need be a biginteger.

...and where it does just what it looks like it should. :O)



The only problem is that... I'd have to... ^{#define mod}

Yeah, there, I said it.

Reasonable for pretty vanities? Or am I just too tired to think straight?

[Dy7SLyTq]

so is mod an operator then? and whats its role? does it perform the mod() function or is it like modulus?

[Duthomhas]

http://en.wikipedia.org/wiki/Modular_arithmetic

[Dy7SLyTq]

wow that is cool

[Disch]

I'm confused on 2 things.

The weird parenthesis placement is the main one. Are you modding b? Or (a+b)?

And secondly... why would mod have to be a macro? Why couldn't you do the same with d = (a + b) % c;?

bigint operator + (int, bigint)
bigint operator % (bigint, int)

Or am I missing something?

[Duthomhas]

The standard mathematical notation is that

    a ≡ b (mod c)

and is read "a is congruent to b mod c".


This kind of stuff is actually useful, though it is most enjoyable when you play with Number Theory.

A common algorithm is to compute

    Y ≡ G^x (mod P)

for cryptographic uses.

Other basic uses include the Chinese Remainder Theorem, etc.


To answer your question, yes, I could write d = (a + b) % c. But with big numbers time is often a factor. There are often ways to compute the answer more quickly than the straightforward method.

By overloading a biginteger () operator to play nice, I can use those functions while allowing a natural (mathematically speaking) syntax.

There are all kinds of ways to apply the optimized algorithms over the standard ones:

d = a.add_mod( b, c );  // single function call

d = a.add( b ).mod( c );  // stack-based delayed eval

d = (a + b).mod( c );  // same

d = a + b (mod c);  // oooh! pretty! 

The whole trick, of course, is that the argument to b() is two lexemes... "mod" and "c", one of which must evaluate to more or less nothing. I know of no other way to do RLM without macros.

Your response is telling me that it is probably a bad idea...

[ne555]

¿So how is a + b(c) actually computing (a+b)(c)?

[Duthomhas]

Stack-based-delayed evaluation. b(c) returns a temp object.

Yeah, I know it was a dumb idea... It sure looked pretty, though.


No matter what you know, the temptation to do something stupid will always come.