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Programming-Idioms

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Idiom #141 Iterate in sequence over two lists

Iterate in sequence over the elements of the list items1 then items2. For each iteration print the element.

from itertools import chain
for x in chain(items1, items2): print(x)
for x in items1 + items2:
    print(x)

;; for side effects
(doseq [x (concat items1 items2)]
   (println x))

void print_seq(const auto& ...xs)
{
 (std::for_each(std::begin(xs), std::end(xs), 
           [](const auto& x) { std::cout << x; }), ...);
}

 std::list xs { "hi", "there", "world" }, ys { "lo" "thar" };

 print_seq(xs, ys);
using System;
using System.Linq;
foreach (var item in items1.Concat(items2))
{
    Console.WriteLine(item);
}
items1.ForEach(Console.WriteLine);
items2.ForEach(Console.WriteLine);
import std.range;
import std.algorithm.iteration;
import std.stdio;
chain(items1, items2).each!writeln;
items1.forEach(print);
items2.forEach(print);
def main(items1, items2) do
  Enum.each(items1, &IO.puts/1)
  Enum.each(items2, &IO.puts/1)
end
items1 ++ items2 |> Enum.each(&IO.puts/1)
print *,items1,items2
for _, v := range items1 {
	fmt.Println(v)
}
for _, v := range items2 {
	fmt.Println(v)
}
mapM_ print $ items1 ++ items2
for (let item of items1) console.log(item)
for (let item of items2) console.log(item)
items1.concat(items2).forEach(console.log)
items1.forEach(System.out::println);
items2.forEach(System.out::println);
import static java.lang.System.out;
for (E e : items1) out.println(e);
for (E e : items2) out.println(e);
import java.util.stream.Stream;
Stream.concat(
	items1.stream(), 
	items2.stream()
).forEach(System.out::println);
(map nil #'print (append items1 items))
# beginner style
foreach( $items1 as $item ) print "$item\n";
foreach( $items2 as $item ) print "$item\n";

# five times faster
print implode("\n", $items1)."\n".implode("\n", $items2)."\n" ;
for i := 0 to items1.Count-1 do writeln(items1[i]);
for i := 0 to items2.Count-1 do writeln(items2[i]);
print for @items1, @items2;
items1.chain(items2).each{|item| puts item}
[items1, items2].each{|ar| ar.each{|item| p item }}
for i in item1.iter().chain(item2.iter()) {
    print!("{} ", i);
}
(map (lambda (x)
       (display x)
       (newline))
     (append items1 items2))
New implementation...
< > ⌨
BBaz 
for x in chain(items1, items2): print(x)
for x in items1 + items2:
    print(x)

;; for side effects
(doseq [x (concat items1 items2)]
   (println x))

void print_seq(const auto& ...xs)
{
 (std::for_each(std::begin(xs), std::end(xs), 
           [](const auto& x) { std::cout << x; }), ...);
}

 std::list xs { "hi", "there", "world" }, ys { "lo" "thar" };

 print_seq(xs, ys);
foreach (var item in items1.Concat(items2))
{
    Console.WriteLine(item);
}
items1.ForEach(Console.WriteLine);
items2.ForEach(Console.WriteLine);
chain(items1, items2).each!writeln;
items1.forEach(print);
items2.forEach(print);
def main(items1, items2) do
  Enum.each(items1, &IO.puts/1)
  Enum.each(items2, &IO.puts/1)
end
items1 ++ items2 |> Enum.each(&IO.puts/1)
print *,items1,items2
for _, v := range items1 {
	fmt.Println(v)
}
for _, v := range items2 {
	fmt.Println(v)
}
mapM_ print $ items1 ++ items2
for (let item of items1) console.log(item)
for (let item of items2) console.log(item)
items1.concat(items2).forEach(console.log)
items1.forEach(System.out::println);
items2.forEach(System.out::println);
for (E e : items1) out.println(e);
for (E e : items2) out.println(e);
Stream.concat(
	items1.stream(), 
	items2.stream()
).forEach(System.out::println);
(map nil #'print (append items1 items))
# beginner style
foreach( $items1 as $item ) print "$item\n";
foreach( $items2 as $item ) print "$item\n";

# five times faster
print implode("\n", $items1)."\n".implode("\n", $items2)."\n" ;
for i := 0 to items1.Count-1 do writeln(items1[i]);
for i := 0 to items2.Count-1 do writeln(items2[i]);
print for @items1, @items2;
items1.chain(items2).each{|item| puts item}
[items1, items2].each{|ar| ar.each{|item| p item }}
for i in item1.iter().chain(item2.iter()) {
    print!("{} ", i);
}
(map (lambda (x)
       (display x)
       (newline))
     (append items1 items2))