depp

-- See http://stackoverflow.com/questions/9967790/instance-in-haskell

instance (Enum x, Enum y) => Enum (x, y) where
  fromEnum (x,y) =
    k^2 + 2*j + if permuted then 1 else 0
      where
        k = max (fromEnum x) (fromEnum y)
        j = min (fromEnum x) (fromEnum y)
        permuted = (fromEnum y) > (fromEnum x)
  toEnum n =

depp

import os, time
def list_modules(path):
    for name in os.listdir(path):
        base, ext = os.path.splitext(name)
        if '.' in base or base.startswith('__'):
            continue
        if ext == '.py':
            yield base
        elif not ext:
            if not os.path.exists(os.path.join(base, '__init__.py')):

depp

/* Copyright 2012 Dietrich Epp <[email protected]> */
#if defined(__GNUC__) && \
    ((__GNUC__ == 4 && __GNUC_MINOR >= 6) || __GNUC__ > 4)
#define HAVE_DPUSH 1
#endif

#include "keycode/keycode.h"
#include "keycode/keytable.h"
#include "sg/audio_system.h"
#include "sg/clock.h"

depp

fn main() {
    let mut x = "hello".chars();
    while (true) {
        match x.next() {
            Some('o') => return,
            Some(c) => println!("Char: {}", c),
            None => return
        }
    }
}

depp

#include <iostream>
#include <vector>

#undef CUSTOM_BUFFER

#if defined CUSTOM_BUFFER
static const int BUFFER_SIZE = 1024 * 16;
#endif

int main()

depp

class A(object):
    def __init__(self):
        print('A')

class B(A):
    def __init__(self):
        print('B')
        super(self.__class__, self).__init__()

class C(B):

depp

run: test.o main.o
	cc -o $@ $^
test.o: test.c
	cc -Wall -Wextra -O3 -c $< -o $@
main.o: main.c
	cc -Wall -Wextra -O2 -c $< -o $@

depp

As you can see, only extern_func() is accessible from other translation units.

$ c++ -std=c++11 -c test.cpp
$ nm -g test.o | c++filt
0000000000000000 T extern_func()
$ nm test.o | c++filt 
0000000000000000 T extern_func()
0000000000000006 t static_func()
000000000000000c t (anonymous namespace)::anonymous_namespace_func()

depp

This is what happens when I run the program.  Note that 2 and 4 appear twice, but 0 and 1 never appear.  The output is non-deterministic, because there is a race condition, the output could be almost anything.

Arg = 2
Arg = 4
Arg = 3
Arg = 5
Arg = 6
Arg = 2
Arg = 7
Arg = 8

depp

Notice how each number appears only once.

The order is still scrambled, but we don't care about the order.

Arg = 1
Arg = 4
Arg = 2
Arg = 6
Arg = 3
Arg = 7