blocks|key|2270494|text|删除最后的System.out.print(n%2B“")；在递归数字调用之后，它返回n的原始值，并再次打印这个值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2270495|它从最深的层次返回，打印1次，直到第3次的调用，再次打印，同时5也被再次打印。|2270496|如果希望它在第一次打印后更新该值，则必须执行n-=2而不是n-2来更新变量n。|2270497|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Remove the final System.out.print(n+" ");
After the recursive numbers call it comes back with the original value for n and prints this again.

It bubbles back from the deepest level where it prints 1 two times up to the call with number 3, which is printed again, and also 5 is printed again.

If you want it to update the value after it is printed the first time you will have to update the variable n by performing n-=2 instead of n-2

blocks|key|17928|text|下面是幕后发生的事情，对于n+==+5|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|17929|numbers(5);
++++if(5+>+0)-->+true+:+
++++++++System.out.print(5+%2B+"+");+//+(1)
++++++++numbers(3);
++++++++%7C+++if(3+>+0)-->+true+:+
++++++++%7C+++++++System.out.print(3+%2B+"+");+//+(2)
++++++++%7C+++++++numbers(1);
++++++++%7C+++++++%7C+++if(1+>+0)-->+true+:+
++++++++%7C+++++++%7C+++++++System.out.print(1+%2B+"+");+//+(3)
++++++++%7C+++++++%7C+++++++numbers(-1);
++++++++%7C+++++++%7C+++++++System.out.print(1+%2B+"+");+//+(4)
++++++++%7C+++++++System.out.print(3+%2B+"+");+//+(5)
++++++++System.out.print(5+%2B+"+");+//+(6)|code-block|syntax|javascript|17930|注意每个数字应该如何被打印两次：|BOLD|17931|System.out.print(n+%2B+"+");+//+print+once
numbers(n-2);
System.out.print(n+%2B+"+");+//+print+twice|17932|entityMap^0|D|6|0|0|6|5|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|U|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|V|8|@$9|W|A|X|B|M]]|D|@]|E|$]]|$1|N|3|O|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|P|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|Q|$]]

Here is what happens behind the scenes, for <code>n == 5</code>:

<pre><code>numbers(5);
 if(5 &gt; 0)--&gt; true : 
 System.out.print(5 + " "); // (1)
 numbers(3);
 | if(3 &gt; 0)--&gt; true : 
 | System.out.print(3 + " "); // (2)
 | numbers(1);
 | | if(1 &gt; 0)--&gt; true : 
 | | System.out.print(1 + " "); // (3)
 | | numbers(-1);
 | | System.out.print(1 + " "); // (4)
 | System.out.print(3 + " "); // (5)
 System.out.print(5 + " "); // (6)
</code></pre>

Notice how each number is supposed to be printed twice:

<pre><code>System.out.print(n + " "); // print once
numbers(n-2);
System.out.print(n + " "); // print twice
</code></pre>

blocks|key|17934|text|++++5+first+system.out+then+number(5-2)
++++%7C
++++---->+3++first+system.out+then+number(5-2)
++++++++++%7C
+++++++++++----->1+first+system.out+then+number(5-2)
+++++++++++++++++(smaller+than+0)+,+returning
+++++++++++++++++%7C
+++++++++++++++++1+second+system.out
+++++++++++++++++%7C
+++++++++++3<----++second+system.out
+++++++++++%7C
5<---------++second+system.out|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|17935|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> 5 first system.out then number(5-2)
 |
 ----&gt; 3 first system.out then number(5-2)
 |
 -----&gt;1 first system.out then number(5-2)
 (smaller than 0) , returning
 |
 1 second system.out
 |
 3&lt;---- second system.out
 |
5&lt;--------- second system.out
</code></pre>

blocks|key|2270589|text|在n=-1之后，递归调用将结束并返回给调用方方法以继续。它发现的第一个命令是打印。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2270590|如果您尝试使用调试器进行调试，您将看到它是多么的合理。|2270591|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

after n=-1, the recursive call will end and return back to the caller method to continue. first commands it found is print.

If you try to debug with the debugger, you will see how rational it is.

blocks|key|2361489|text|Java将方法调用放到堆栈中，然后打印第5,3,1次方法调用，然后恢复对其余调用的执行，然后从堆栈中打印1,3,5次(堆栈中的最后一次调用先被拾取)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2361490|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Java maintain method call into stack and as first 5,3,1 gets printed after which method call resume execution for remaining calls and prints 1,3,5 from stack(last call in stack get's picked up first).

blocks|key|2270655|text|遵循递归堆栈。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2270656|具有和输出流的堆栈表示：|2270657|+++++++++++++------------------------->
+++++++++++++5+++++++++++3++++++++++++1
num(5)-->++num(3)+-->+num(1)+-->+num(-1)+-->+void
+++++++++++++5+++++++++++3++++++++++++1
+++++++++++++<-------------------------|code-block|syntax|javascript|2270658|现在，由于数字(-1)不满足if条件，程序控制从if块中出来，并返回void。现在开始弹出堆栈(递归)：5，3，1，15，5，这是您得到的输出，也是预期的输出。|2270659|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Follow the recursion stack.

stack representation with and output flow:

<pre><code> -------------------------&gt;
 5 3 1
num(5)--&gt; num(3) --&gt; num(1) --&gt; num(-1) --&gt; void
 5 3 1
 &lt;-------------------------
</code></pre>

Now since numbers(-1) does not satisfy the if condition, the program control comes out of the if block, and returns void. 
Start popping out the stack now (recursion):
5 3 1 1 3 5. That's what the output you get, and is expected.

When I run this piece of code for n=5 the output I get is "5 3 1 1 3 5"
I get the 5 3 1 part but after that n=-1 but when I run the code with a debugger it when n=-1 it goes to the line after <code>numbers(n-2);</code>i.e <code>System.out.prt(n+ "");</code> even though that statement is contained in the if block.

Why does this happen? 

<pre><code>public void numbers(int n)
{
 if(n&gt;0)
 {
 System.out.print(n+" ");
 numbers(n-2);
 System.out.print(n+" ");
 }
}
</code></pre>

TLDR : when n=-1 <code>System.out.prt(n+ "");</code>even though it is within the if block which only runs when n>0.

Any help would be much appreciated. Thanks in advance!

How does this recursive function get to this output?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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当我为n=5运行这段代码时，我得到的输出是"5 3 1 1 3 5“，我得到了5 3 1部分，但在那之后，n=-1，但是当我使用调试器运行代码时，当n=-1时，它会转到numbers(n-2);i.e System.out.prt(n+ "");之后的行，即使该语句包含在if块中。这一切为什么要发生？public vo...

问这个递归函数是如何得到这个输出的？
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问这个递归函数是如何得到这个输出的？EN