Java 显示从 1 到 100 的数字，没有循环或条件
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Display numbers from 1 to 100 without loops or conditions
提问by Thunderhashy
Is there a way to print numbers from 1 to 100 without using any loops or conditions like "if"? We can easily do using recursion but that again has an if condition. Is there a way to do without using "if" as well? Also no repetitive print statements, or a single print statement containing all the numbers from 1 to 100.
有没有办法在不使用任何循环或条件（如“if”）的情况下打印从 1 到 100 的数字？我们可以很容易地使用递归，但这也有一个 if 条件。有没有办法不使用“if”？也没有重复的打印语句，或包含从 1 到 100 的所有数字的单个打印语句。
A solution in Java is preferable.
Java 中的解决方案是可取的。
采纳答案by Yacoby
Pseudo code. Uses an array to force an exception after 100 elements which is caught and does nothing.
伪代码。使用数组在 100 个被捕获但不执行任何操作的元素后强制执行异常。
function r(array a, int index){
a[index] = a[index-1]+1
print a[index]
r(a, index+1)
}
try{
array a;
a.resize(101)
r(a, 1)
}catch(OutOfBoundsException){
}
EDIT
Java code:
编辑
Java代码：
public void printTo100(){
int[] array = new int[101];
try{
printToArrayLimit(array, 1);
}catch(ArrayIndexOutOfBoundsException e){
}
}
public void printToArrayLimit(int[] array, int index){
array[index] = array[index-1]+1;
System.out.println(array[index]);
printToArrayLimit(array, index+1);
}
回答by truppo
Sure there is:
当然有：
System.out.println(1);
System.out.println(2);
System.out.println(3);
System.out.println(4);
System.out.println(5);
System.out.println(6);
System.out.println(7);
System.out.println(8);
System.out.println(9);
System.out.println(10);
System.out.println(11);
System.out.println(12);
System.out.println(13);
System.out.println(14);
System.out.println(15);
System.out.println(16);
System.out.println(17);
System.out.println(18);
System.out.println(19);
System.out.println(20);
System.out.println(21);
System.out.println(22);
System.out.println(23);
System.out.println(24);
System.out.println(25);
System.out.println(26);
System.out.println(27);
System.out.println(28);
System.out.println(29);
System.out.println(30);
System.out.println(31);
System.out.println(32);
System.out.println(33);
System.out.println(34);
System.out.println(35);
System.out.println(36);
System.out.println(37);
System.out.println(38);
System.out.println(39);
System.out.println(40);
System.out.println(41);
System.out.println(42);
System.out.println(43);
System.out.println(44);
System.out.println(45);
System.out.println(46);
System.out.println(47);
System.out.println(48);
System.out.println(49);
System.out.println(50);
System.out.println(51);
System.out.println(52);
System.out.println(53);
System.out.println(54);
System.out.println(55);
System.out.println(56);
System.out.println(57);
System.out.println(58);
System.out.println(59);
System.out.println(60);
System.out.println(61);
System.out.println(62);
System.out.println(63);
System.out.println(64);
System.out.println(65);
System.out.println(66);
System.out.println(67);
System.out.println(68);
System.out.println(69);
System.out.println(70);
System.out.println(71);
System.out.println(72);
System.out.println(73);
System.out.println(74);
System.out.println(75);
System.out.println(76);
System.out.println(77);
System.out.println(78);
System.out.println(79);
System.out.println(80);
System.out.println(81);
System.out.println(82);
System.out.println(83);
System.out.println(84);
System.out.println(85);
System.out.println(86);
System.out.println(87);
System.out.println(88);
System.out.println(89);
System.out.println(90);
System.out.println(91);
System.out.println(92);
System.out.println(93);
System.out.println(94);
System.out.println(95);
System.out.println(96);
System.out.println(97);
System.out.println(98);
System.out.println(99);
System.out.println(100);
回答by fastcodejava
Try :
尝试 ：
int i = 1;
System.out.println(i++);
System.out.println(i++);
System.out.println(i++);
....
回答by LBushkin
Yes, it's possible, but it's terrible.There's any number of ways that will use either recursion or nested type creation, with exception handling for flow control. This has no real-world application IMO and should be avoided in real code at all costs.
是的，这是可能的，但这很可怕。有许多方法可以使用递归或嵌套类型创建，以及用于流控制的异常处理。这在 IMO 中没有实际应用程序，应该不惜一切代价在实际代码中避免。
Here's an example that uses recursive type instantiation with exception handling to control termination. It print the number is descending order, but that would be trivial to change to ascending, by just subtracting 99 (or whatever constant) from the value being printed.
这是一个使用递归类型实例化和异常处理来控制终止的示例。它以降序方式打印数字，但只需从正在打印的值中减去 99（或任何常数）即可更改为升序。
class PrintVal
{
// called with a list containing as many items as you want, less one...
public PrintVal( List<int> items )
{
System.out.println(items.size()+1); // print the size of the list
try {
items.remove( items.size()-1 ); // will throw when items is empty
new PrintVal( items );
}
catch( Exception ) { /* swallow and terminate */ }
}
}
// setup and invocation that performs the output
ArrayList<int> strList = new ArrayList<int>( new int[99] );
PrintVal instance = new PrintVal( strList ); // all output happens here
回答by Adam Pope
Check out the Divide + Conquer answer from the C# thread. It's evil, but brilliant:
查看 C# 线程中的 Divide + Conquer 答案。这是邪恶的，但很精彩：
How to print 1 to 100 without any looping using C#
Here is the Java version:
这是Java版本：
public class Application {
public static void main(String[] args) {
Print64Numbers();
Print32Numbers();
Print4Numbers();
}
private static int currentNumber = 0;
private static void Print1Number() { System.out.println(++currentNumber); }
private static void Print2Numbers() { Print1Number(); Print1Number(); }
private static void Print4Numbers() { Print2Numbers(); Print2Numbers(); }
private static void Print8Numbers() { Print4Numbers(); Print4Numbers(); }
private static void Print16Numbers() { Print8Numbers(); Print8Numbers(); }
private static void Print32Numbers() { Print16Numbers(); Print16Numbers(); }
private static void Print64Numbers() { Print32Numbers(); Print32Numbers(); }
}
回答by b_erb
Implement a recursive call incrementing and printing the number. Configure your VM to run out of stack after 100 calls. No conditions, no loops. cough;-)
实现递归调用递增并打印数字。将 VM 配置为在 100 次调用后耗尽堆栈。没有条件，没有循环。咳嗽;-)
回答by Sajad Bahmani
without any loop and condition :
没有任何循环和条件：
public static void recfunc(int a[], int i)
{
System.out.println(i);
int s = a[i];
recfunc(a, i + 1);
}
public static void main(String[] args)
{
int[] a = new int[100];
try
{
recfunc(a, 1);
}
catch (Exception e)
{
}
}
with recursion and without if I think use "?" for conditioning :
有递归，没有如果我认为使用“？” 用于调理：
public static int recfunc(int i)
{
System.out.println(i);
return (i < 100) ? recfunc(i + 1) : 0;
}
public static void main(String[] args)
{
recfunc(1);
}
回答by Arne Deutsch
Abuse an exception to serve as a condition.
滥用异常作为条件。
public class Main {
private static int[] stopper = new int[100];
public static void main(String[] args) {
try {
print(1);
} catch(ArrayIndexOutOfBoundsException e) {
// abuse of try catch
}
}
private static void print(int i) {
System.out.println(i);
stopper[i] = i;
print(i + 1);
}
}
回答by S.Lott
Here's a hint that be helpful.
这是一个有用的提示。
The assertstatement isn't the forbidden ifstatement.
该assert声明是不被禁止的if声明。
回答by z -
DO NOT DO THIS UNDER ANY SANE CIRCUMSTANCES!
请勿在任何正常情况下执行此操作！
public class Fail {
public void thisFails(int x){
System.out.println(x);
Integer[] bigArray = new Integer[9450];
thisFails(x+1);
}
public static void main(String[] args) {
Fail failure = new Fail();
failure.thisFails(1);
}
}
When this is ran using 1m of heap space (java -Xmx1m Fail) it will run out of heap at the 100th recursion.
当它使用 1m 的堆空间（java -Xmx1m Fail）运行时，它将在第 100 次递归时用完堆。
...
...
I will now go wash my hands.
我现在去洗手。
