The Nineteenth Byte

DJMcMayhem

6:00 PM

C golfing question: which compilers allow f(a,b) in place of int f(int a, int b)?

quartata

@DJMcMayhem GCC is fine with K&R for the most part.

DJMcMayhem

You're right, GCC seems very forgiving: Try it online!

At least, for C. C++ less so

quartata

Yeah, most of the weird weak typing possible in C is not allowed in any C++ compiler.

Part of the standard.

Kritixi Lithos

@DJMcMayhem Try it online!

Erik the Outgolfer

Actually, it seems TCC is even more forgiving, given that you do not even need return statements in functions, and that it's based of GCC.

Kritixi Lithos

6:10 PM

@EriktheOutgolfer You don't need returns in GCC under special circumstances: tio.run/nexus/…

2

DJMcMayhem

Woah. That's a huge golfing tip. What are those circumstances?

Erik the Outgolfer

Huh? I'd swear this didn't work earlier that I had tried.

Anonymous

@DJMcMayhem The implicit return type is the result of the last expression (usually/sometimes)

Erik the Outgolfer

@DJMcMayhem Yeah, it even beats preprocessor solutions.

Anonymous

Err, rather, it's stored in the first argument

Kritixi Lithos

6:12 PM

7

A: Tips for golfing in C

Assign instead of return. This is not really standard C, but works with every compiler and CPU that I know of: int sqr(int a){return a*a;} has the same effect as: int sqr(int a){a*=a;} Because the first argument is stored into the same CPU register as the return value. X-Macros Another u...

DJMcMayhem

@Mego Then why doesn't this work: Try it online!

Anonymous

You have to store it in a

DJMcMayhem

Nevermind

Then why does this work? Try it online!

Anonymous

That's weird

Kritixi Lithos

@DJMcMayhem Because you are assigning something to something

quartata

6:14 PM

That might be a GNU C extension.

Anonymous

I always thought it was the first arg

quartata

Oh wait nevermind

I misread it

Kritixi Lithos

Apparently when I do a+=2 (or even b+=2) it returns 0

DJMcMayhem

Also, why does this work Try it online! but this doesn't: Try it online!

Anonymous

But when you do a+=a, it works

DJMcMayhem

6:15 PM

(I'm filled with questions today)

Anonymous

It must be because a temporary copy of a is pushed on the stack, and that is implicitly popped and returned

Anonymous

a+=2 probably gets optimized

Kritixi Lithos

@DJMcMayhem But this works: Try it online!

Anonymous

@DJMcMayhem Second one is a constant function that doesn't depend on the args, so it gets optimized

DJMcMayhem

@KritixiLithos woooooaaah o_O

Kritixi Lithos

6:17 PM

It's like magiC

2

New Main Posts

0

Q: Find Recursively Prime Primes

The Recursively Prime Primes is are sequence of primes such that p(1) = 2 p(n) = the p(n-1)th prime Here is an example of how one might calculate the 4th Recursively Prime Prime. p(4) = the p(3)th prime p(3) = the p(2)th prime p(2) = the p(1)th prime p(1) = 2 p(2) = the 2nd prime p(2) = 3 p(...

DJMcMayhem

I regularly use C, I'm obsessed with code-golf, how have I never known this before?

Kritixi Lithos

@DJMcMayhem That's how I felt when I saw this tip for golfing in Processing being used in an answer.

DJMcMayhem

It's even cooler when it's a language you use regularly. Good vim answers blow my mind

Doorknob

@KritixiLithos wait, what?

That's so bizarre

Riley

6:31 PM

@Doorknob I think eax is just a convenient register, so the compiler puts whatever it can there, and that happens to be where it looks for a return value later.

Dennis

@Doorknob Not really. gcc doesn't try very hard to optimize by default, but it still recognizes int c = 42; as useless and optimizes it out. With -O1, the int c = a*b; version also stops working.

Doorknob

I mostly meant the whole "return without return" thing in general

Kritixi Lithos

@Dennis But why is int c = 42; useless but c = 42 + (b=0); not?

DJMcMayhem

This is shorter: Try it online!

@KritixiLithos because it forces the compiler to evaluate b?

Dennis

@KritixiLithos Both are useless, but the first is easier to recognize.

Anonymous

6:35 PM

@KritixiLithos Because it references one of the args. Since gcc isn't trying too hard to optimize at default optimization, it doesn't expend the effort to realize that is the same.

betseg

is youtube down

New Sandboxed Posts

0

A: Sandbox for Proposed Challenges

Solve The Halting Problem Write a program that will take the text of another program as input. Your program will examine the code and output a result that the program either halts or will keep running forever. That's impossible? Nonsense! To show how wrong you are, I've written this program tha...

Kritixi Lithos

@DJMcMayhem But int c = 42+b-b; doesn't work...

@betseg Not for me

betseg

gives 500 for me

Business Cat

downforeveryoneorjustme.com/youtube.com

betseg

6:37 PM

> It's not just you! youtube.com looks down from here.

DJMcMayhem

@KritixiLithos super weird

Dennis

@KritixiLithos That's also an easy optimization, as -b+b is superfluous in all well-defined situations. This isn't the case for /b*b; you have to know the value of b to know the result.

Business Cat

Other similar sites say it's up though.

Dennis

Oh, it was *b/b, not /b*b.

Pavel

Is b an int or a double?

DJMcMayhem

6:39 PM

Int

@Dennis because it might be zero or out of bounds?

Pavel

3/2*2==2

Dennis

@DJMcMayhem Yeah, it looks like gcc has reservations about this, even with -Ofast. Division by 0 is still UB though, so the compiler could technically optimize out *b/b out if it wanted to.

So is signed integer overflow.