Java VS Ruby Returns
JAVAの教科書にあったソース
class Dog {
int weight = 50;
static {
System.out.println(“static inr”);
}
{
System.out.print(“inr “);
}
Dog() {
System.out.println(“ctr”);
}
void meth() {
System.out.println(weight);
}}
class SampleA {
public static void main(String[] args){
Dog x = new Dog();
Dog y = new Dog();
x.meth();
}
}
Rubyに翻訳
class Dog
def initialize
@weight = 50
p “inr ctr”
end
def meth
p @weight
end
endp “static inr”
x = Dog.new
y = Dog.new
x.meth
もういっちょJAVA
import java.util.*;
class SampleB {
public static void main(String[] args){
int[] ds = {1, 2, 3, 4, 5};
for(int d : ds)
System.out.println(d);
System.out.println();
int[][] nns = {{1, 2, 3}, {10, 20, 30}, {100, 200, 300}};
for(int[] ns : nns){
for(int n : ns){
System.out.println(n);
}
System.out.println();
}
List<String> ss = new ArrayList<String>();
ss.add(“abc”);
ss.add(“def”);
ss.add(“ghi”);
for(String s : ss)
System.out.println(s);
}
}
もういっちょRuby
class SampleB
def main
ds = [1, 2, 3, 4, 5]
for d in ds
p d
end
p “”
nns = [[1, 2, 3], [10, 20, 30], [100, 200, 300]]
for ns in nns
for n in ns
p n
end
p “”
end
ss = []
ss.push(“abc”)
ss.push(“def”)
ss.push(“ghi”)
for s in ss
p s
end
end
endx = SampleB.new
x.main
コメントを残す