blocks|key|1536929|text|这段代码使用C%2B%2B，这就是int+&语法的意思。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1536930|引用基本上是指针的句法糖。因此，在调用f(p,+p)时，函数参数x是main()中对p的引用，而c只是调用时值的副本。|1536931|这意味着f可以通过引用在main()中更改p的值。因为f递归地调用自己，将相同的引用传递给自己，所以它始终是对main()中的p的引用。|1536932|为了追踪回溯，我建议在f()内部的战略位置添加登录打印。|1536933|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Reference_%2528C%2B%2B%2529|1|http://en.wikipedia.org/wiki/Syntactic_sugar^0|D|5|6|3|0|0|J|7|W|1|Y|6|16|1|1C|1|9|3|1|0|4|1|C|6|L|1|R|1|1J|6|1R|1|0|B|3|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@$9|1C|A|1D|1|1E]]|E|$]]|$1|H|3|I|5|6|7|1F|8|@$9|1G|A|1H|B|C]|$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]|$9|1Q|A|1R|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|1S|8|@$9|1T|A|1U|B|C]]|D|@]|E|$]]|$1|L|3|-4|5|6|7|1V|8|@]|D|@]|E|$]]]|M|$N|$5|O|P|Q|E|$R|S]]|T|$5|O|P|Q|E|$R|U]]]]

This code uses a <a href="http://en.wikipedia.org/wiki/Reference_%28C++%29" rel="nofollow">C++ reference</a>, which is what the <code>int &amp;</code> syntax means.

A reference is, basically, <a href="http://en.wikipedia.org/wiki/Syntactic_sugar" rel="nofollow">syntactic sugar</a> for a pointer. So when you call <code>f(p, p)</code>, the function argument <code>x</code> is a reference to <code>p</code> in <code>main()</code>, while <code>c</code> is merely a copy of the value at the time of the call.

This means <code>f</code> can change the value of <code>p</code> in <code>main()</code>, through the reference. Since <code>f</code> calls itself recursively, passing the same reference toitself, it's always a reference to the <code>p</code> in <code>main()</code>.

To track the recusion, I would suggest adding logging print-outs at strategic places inside <code>f()</code>.

blocks|key|2327979|text|第一个是创建一个reference，它更像是一个更干净的语法，而不是一个指针。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2327980|您可以使用它在函数调用中更改变量的内容。|2327981|尽管您必须注意，您发布的代码是无效的C%2B%2B。C%2B%2B不允许没有返回类型的main。|2327982|entityMap^0|8|9|0|0|Z|4|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|K|$]]

The first creates a <code>reference</code> which is more like a cleaner syntax instead of a pointer.

You can use it to change contents of variables among calls to function. 

Although you must note, the code you have posted is not valid C++. C++ does not allow <code>main</code> without a return type.

blocks|key|2327987|text|更简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2327988|int+f(int+&a){
++a+=+5;
}
int+x+=+0;
f(x);
//now+x+equals+5

int+f2(int+b){
+b+=+5;
}
int+y+=+0;
f2(y);
//y+still+equals+0|code-block|syntax|javascript|2327989|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

More simple:

<pre><code>int f(int &amp;a){
 a = 5;
}
int x = 0;
f(x);
//now x equals 5

int f2(int b){
 b = 5;
}
int y = 0;
f2(y);
//y still equals 0
</code></pre>

blocks|key|1536955|text|很明显，问题中的参考部分是为了分散你的注意力，它只是x指向p的地址，但是x本身第一次包含了值5，但是递归运行了4次，所以在堆栈中，它就像--x当值6(堆栈底部)，times值7，times值是8，当值为9(堆栈顶部)时，顶部为9，但正如我们所知，在递归中，第一个循环执行发生，然后求值，所以当x的值更改为9时，堆栈的所有四个块都更改为9，因为它们指向相同的‘x’.所以结果是9*9*9*9=6561+..as就这样简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1536956|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|$]]

Actaully it is like this the reference part in the question was to distract u it is just x pointed to the address of the the p but the 'x' itself contained the value 5 at the first time but the recurrsion ran for 4 times so in the stack it was like ----[x when value 6(bottom of the stack), xwhen value 7, xwhen value was 8 , x when value is 9(top of the stack)] ...at the top it was 9 but as we know that in recurrsion first loop execution takes place and then the evaluation so when the value of x changed to 9 then all the four block of the stack changed to 9 since they pointed to the same 'x'....so it came out to be 9*9*9*9=6561 ..as simple as that.

blocks|key|2328012|text|&是引用:它意味着它必须是对象的引用(地址)；|type|unordered-list-item|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2328013|*是一个指针:它指向一个对象。|2328014|区别在于"&x“不能是"NULL"，而&x可以是NULL。|unstyled|2328015|这两个原型是不同的。|2328016|int+f(int+&+,int+);
int+f(int+*+,int+);+//are+different+function+prototypes|code-block|syntax|javascript|2328017|这是个好问题。|2328018|entityMap^0|0|1|0|0|1|0|5|2|C|4|J|2|O|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|K|3|L|5|J|7|1A|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|1B|8|@]|D|@]|E|$P|Q]]|$1|R|3|S|5|J|7|1C|8|@]|D|@]|E|$]]|$1|T|3|-4|5|J|7|1D|8|@]|D|@]|E|$]]]|U|$]]

<ul>
<li><code>&amp;</code> is reference: it means it must be an object's reference (address);</li>
<li><code>*</code> is a pointer: it points to a object.</li>
</ul>

The difference is that "<code>&amp;x</code>" can not be "<code>NULL</code>", but 
<code>&amp;x</code> can be <code>NULL</code>.

These two prototype are different.

<pre><code>int f(int &amp; ,int );
int f(int * ,int ); //are different function prototypes
</code></pre>

This is a good question.

blocks|key|2328048|text|该代码正在计算(2P1)%5E(P-1)，尝试各种P值以查找模式。在这里，P=5；所以它产生9%5E4+=+6561。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2328049|پویاپاکاریان|2328050|2328051|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|H|8|@]|9|@]|A|$]]|$1|D|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|E|3|-4|5|6|7|J|8|@]|9|@]|A|$]]]|F|$]]

The code is calculating (2P-1)^(P-1)
Try various values for P to find the pattern.
Here, P=5; so it yields 9^4 = 6561.
پویا پاکاریان
پویاپاکاریان

<pre><code>#include&lt;stdio.h&gt;
#include&lt;conio.h&gt;

int f(int &amp; ,int );//function prototype

main()
{
 int x,p=5; //p is initialized to 5
 x=f(p,p);
 printf("\n Value is : %d",x);//print the value of x
 getch();
}

int f (int &amp; x, int c)
{
c=c-1;
if (c==0) return 1;
x=x+1;
return f(x,c) * x; //recursion
}
</code></pre>

output : 6561

can anyone explain me the flow the program
This question is from gate i couldn't understand it. It seems that the function is called with value of p = 5. It is catched in the function f by int &amp;x the problem is here. Is the value i.e 5 is stored in x or in address of x.

difference between int f(int &x) int f(int x)

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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#include<stdio.h>#include<conio.h>int f(int & ,int );//function prototypemain(){ int x,p=5; //p is initialized to 5 x=f(p,p); printf("\n Value is : %d",x);//...

问int f(int &x) int f(int x)
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问int f(int &x) int f(int x)EN