entityMap|blocks|key|fqm58|text|f(5)+=+f(4)++++++++++++++++++++++++%2B+f(3)
+++++=+f(3)+++++++++++%2B+f(2)+++++++%2B+f(2)++++++++%2B+f(1)
+++++=+f(2)+++++%2Bf(1)+%2B+f(1)+%2B+f(0)+%2B+f(1)+%2B+f(0)+%2B+2
+++++=+f(1)%2Bf(0)%2B+2+++%2B+2++++%2B+2++++%2B+2++++%2B+2++++%2B+2
+++++=+2+++%2B+2++%2B+2+++%2B+2++++%2B+2++++%2B+2++++%2B+2++++%2B+2
+++++=+16|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|bhpl0|希望上面的解释能解释16值|unstyled|5hpnm^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|J|9|@]|A|@]|B|$]]|$2|H|4|-4|6|G|8|K|9|@]|A|@]|B|$]]]]

<pre><code>f(5) = f(4) + f(3)
 = f(3) + f(2) + f(2) + f(1)
 = f(2) +f(1) + f(1) + f(0) + f(1) + f(0) + 2
 = f(1)+f(0)+ 2 + 2 + 2 + 2 + 2 + 2
 = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2
 = 16
</code></pre>

Hope the above interpretation will explain the value 16

entityMap|blocks|key|7c4u0|text|f(0)=2
f(1)=2
f(2)=f(1)%2Bf(0)=2%2B2=4
f(3)=f(2)%2Bf(1)=4%2B2=6
f(4)=f(3)%2Bf(2)=6%2B4=10
f(5)=f(4)%2Bf(3)=10%2B6=16|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5i7mv|unstyled^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|G|9|@]|A|@]|B|$C|D]]|$2|E|4|-4|6|F|8|H|9|@]|A|@]|B|$]]]]

entityMap|blocks|key|egu76|text|下面是递归函数的可视化表示，显示了每次迭代的结果|type|unstyled|depth|inlineStyleRanges|entityRanges|data|fofug|++++++++++++++++++++++f(5)++++++++++++++++++++++++++1st
++++++++++++++++++++/+++++++\
+++++++++++++++++++/+++++++++\
++++++++++++++++++/+++++++++++\
+++++++++++++++++/+++++++++++++\
++++++++++++++++/+++++++++++++++\
+++++++++++++++/+++++++++++++++++\
++++++++++++f(4)+++++++%2B++++++++f(3)++++++++++++++++2nd
++++++++++/+++++\+++++++++++++++/++\
+++++++++/+++++++\+++++++++++++/++++\
++++++f(3)++%2B+++f(2)+++%2B++++f(2)++%2Bf(1)+++++++++++++3rd
++++++/++\++++++/++\++++++++/++\+++++%7C
+++f(2)%2Bf(1)%2Bf(1)%2Bf(0)+%2B+f(1)%2Bf(0)%2B++2++++++++++++++4th
+++/++\+++%7C++++%7C+%2B++%7C++++++%7C++++%7C++++%7C
f(1)%2Bf(0)%2B2+%2B++2+%2B++2++%2B+++2+%2B++2+%2B++2++++++++++++++5th
++%7C++++%7C++%7C++++%7C++++%7C++++++%7C++++%7C++++%7C
++2+%2B++2+%2B2+%2B++2+%2B++2++%2B+++2+%2B++2+%2B++2+=+16++++++++Result|code-block|syntax|javascript|8kvnd^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

Here is a visual representation of the recursive function, showing the return on each iteration

<pre><code> f(5) 1st
 / \
 / \
 / \
 / \
 / \
 / \
 f(4) + f(3) 2nd
 / \ / \
 / \ / \
 f(3) + f(2) + f(2) +f(1) 3rd
 / \ / \ / \ |
 f(2)+f(1)+f(1)+f(0) + f(1)+f(0)+ 2 4th
 / \ | | + | | | |
f(1)+f(0)+2 + 2 + 2 + 2 + 2 + 2 5th
 | | | | | | | |
 2 + 2 +2 + 2 + 2 + 2 + 2 + 2 = 16 Result
</code></pre>

entityMap|blocks|key|2bo1j|text|下面是一个更容易阅读的计算明细：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|ekdt4|f(5+-+1)+%2B+f(5+-+2)

(f(4+-+1)+%2B+f(4+-+2))+%2B+(f(3+-+1)+%2B+f(3+-+2))

((f(3+-+1)+%2B+f(3+-+2))+%2B+(f(2+-+1)+%2B+f(2+-+2)))+%2B+((f(2+-+1)+%2B+f(2+-+2))+%2B+2)

Above+simplifies+to:
((f(2)+%2B+2)+%2B+(2+%2B+2))+%2B+((2+%2B+2)+%2B+2)

Then+to:
((f(2)+%2B+2)+%2B+4)+%2B+(6)

Then+remove+extra+parenthesis:
f(2)+%2B+2+%2B+4+%2B+6

Then+simplify:
f(2)+%2B+12

Continuing:
f(2+-+1)+%2B+f(2+-+2)+%2B+12

Simplify:
2+%2B+2+%2B+12+=+16|code-block|syntax|javascript|56k93^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

Here's a more easy to read break-down of the calculations:

<pre><code>f(5 - 1) + f(5 - 2)

(f(4 - 1) + f(4 - 2)) + (f(3 - 1) + f(3 - 2))

((f(3 - 1) + f(3 - 2)) + (f(2 - 1) + f(2 - 2))) + ((f(2 - 1) + f(2 - 2)) + 2)

Above simplifies to:
((f(2) + 2) + (2 + 2)) + ((2 + 2) + 2)

Then to:
((f(2) + 2) + 4) + (6)

Then remove extra parenthesis:
f(2) + 2 + 4 + 6

Then simplify:
f(2) + 12

Continuing:
f(2 - 1) + f(2 - 2) + 12

Simplify:
2 + 2 + 12 = 16
</code></pre>

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://3v4l.org/1Ob4A|blocks|key|dq83o|text|每当您的递归函数被调用并传递一个大于1的值时，您就会得到一个"fork“(或随后的递归调用)。|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|bhek1|精简语法：(Demo)|5hs42|function+f($n){
++return+$n<2+?+2+:+f($n-1)%2Bf($n-2);
}|code-block|syntax|javascript|dqdue|当为f(1)时，没有fork。返回2|ec11f|当为f(2)时，只有1个fork。返回4+(LHS=2+RHS=2)|jjbm|当为f(3)时，在最初的fork之后，会有一个后续的fork+@+LHS分支。返回6+(LHS=4+RHS=2)|45lcc|当f(4)时，在初始分支之后，有3个分支:+2xLHS和1xRHS。返回10+(LHS=6+RHS=4)|c4q2h|当f(5)时，在初始分支之后，有6个分支:+4xLHS和2x$RHS。返回16+(LHS=10+RHS=6)|fmakm^0|I|1|0|6|4|0|0|0|2|4|H|1|0|2|4|J|1|0|2|4|15|1|0|1|4|10|2|0|1|4|11|2|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|14|G|@$H|15|I|16|J|K]]|L|@]|6|$]]|$A|M|C|N|2|E|F|17|G|@]|L|@$H|18|I|19|A|1A]]|6|$]]|$A|O|C|P|2|Q|F|1B|G|@]|L|@]|6|$R|S]]|$A|T|C|U|2|E|F|1C|G|@$H|1D|I|1E|J|K]|$H|1F|I|1G|J|K]]|L|@]|6|$]]|$A|V|C|W|2|E|F|1H|G|@$H|1I|I|1J|J|K]|$H|1K|I|1L|J|K]]|L|@]|6|$]]|$A|X|C|Y|2|E|F|1M|G|@$H|1N|I|1O|J|K]|$H|1P|I|1Q|J|K]]|L|@]|6|$]]|$A|Z|C|10|2|E|F|1R|G|@$H|1S|I|1T|J|K]|$H|1U|I|1V|J|K]]|L|@]|6|$]]|$A|11|C|12|2|E|F|1W|G|@$H|1X|I|1Y|J|K]|$H|1Z|I|20|J|K]]|L|@]|6|$]]|$A|13|C|-4|2|E|F|21|G|@]|L|@]|6|$]]]]

Every time your recursive function is called and passed a value greater than <code>1</code>, you get a "fork" (or a subsequent recursion call).

Condensed syntax: (<a href="https://3v4l.org/1Ob4A" rel="nofollow noreferrer">Demo</a>)

<pre><code>function f($n){
 return $n&lt;2 ? 2 : f($n-1)+f($n-2);
}
</code></pre>

When <code>f(1)</code>, there is no fork. return <code>2</code>

When <code>f(2)</code>, there is only 1 fork. return <code>4</code> (LHS=2 RHS=2)

When <code>f(3)</code>, after an initial fork, there is a subsequent fork @ LHS branch. return <code>6</code> (LHS=4 RHS=2)

When <code>f(4)</code>, after an initial fork, there are 3 forks: 2xLHS &amp; 1xRHS. return <code>10</code> (LHS=6 RHS=4)

When <code>f(5)</code>, after an initial fork, there are 6 forks: 4xLHS &amp; 2x$RHS. return <code>16</code> (LHS=10 RHS=6)

<pre><code>function f($n)
{
 if ($n == 0 || $n == 1)
 return 2;
 return f($n-1) + f($n-2);
}

echo f(5);
</code></pre>

This gives me <code>16</code> but I have no idea why.

I know it goes something like this: 

<pre><code>return f(4) + f(3)
return f(4) + f(3)+ f(3) + f(1)
</code></pre>

But now since a value of <code>n</code> is <code>1</code>, it should return <code>2</code>.

So it's essentially

<pre><code>return f(4) + f(3)+ f(2)
</code></pre>

Which should be <code>9</code>, but it gives me <code>16</code>.

Can anyone explain how please?

depicting php code functions/recursion; grade 11 functions/recursions

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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 function f($n){   if ($n == 0 || $n == 1)      return 2;   return  f($n-1) + f($n-2);}echo f(5); 这给了我16，但我不知道为什么。 我知道大概是这样的： return f(4) + f(3)return f(4) + f(3)+ f(3) + f(1) 但是现在因为n的值是1，所以它应该返回2。 所以

问描述php代码函数/递归；11级函数/递归
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回答 5

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