blocks|key|1997354|text|total是一个局部变量。一旦myFunction处理完一个数据，它就会被销毁。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1997355|拥有状态的典型方法是使myFunction成为function+object+(一个重载()操作符的结构)。|1997356|最糟糕的方法是让total成为一个全局变量。|1997357|在您的情况下，我建议您使用the+accumulate+function+(假设cout+<<+total只是用于调试)。|1997358|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Function_object#In_C_and_C.2B.2B|1|http://www.sgi.com/tech/stl/accumulate.html^0|0|5|F|A|0|B|A|18|2|N|F|0|0|8|5|0|H|A|14|D|D|N|1|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|H|3|I|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@$9|1G|A|1H|1|1I]]|E|$]]|$1|L|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|M|$N|$5|O|P|Q|E|$R|S]]|T|$5|O|P|Q|E|$R|U]]]]

The <code>total</code> is a local variable. It will be destroyed once the <code>myFunction</code> finished processing one data.

The typical way to have a state is to make <code>myFunction</code> a <a href="http://en.wikipedia.org/wiki/Function_object#In_C_and_C.2B.2B" rel="nofollow">function object</a> (a struct that overloads the <code>()</code> operator). 
The dirty way is to make <code>total</code> a global variable. 
In your case I'd recommend you use <a href="http://www.sgi.com/tech/stl/accumulate.html" rel="nofollow">the <code>accumulate</code> function</a> instead (assuming that <code>cout &lt;&lt; total</code> is just for debugging).

blocks|key|3461457|text|您的total在每个函数调用中都是本地的。它在每次迭代时都被初始化为0。您可以将其声明为全局的(或打包到一个参数中，等等)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3461458|entityMap^0|2|5|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Your <code>total</code> is local at each function call. It's initialized with 0 at every iteration. You could just declare it global (or pack into a parameter, etc.)

blocks|key|1250047|text|你确实需要一个函数器来存储迭代之间的状态：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1250048|struct+Sum
{
++++Sum(int&+v):++value(v)+{}
++++void+operator()(int+data)+const+{+value+%2B=+data;}

++++int&+value;
};

int+main()
{
++++int+array[]+=+{+1,+2,+3,+4,+5,+6,+7,+8,+9,+10+};

++++int+total+=+0;
++++std::for_each(+array,+array%2B10,+Sum(total));

++++std::cout+<<+total+<<+std::endl;
}|code-block|syntax|javascript|1250049|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|$]]

You really need a functor to store state between iterations:

<pre><code>struct Sum
{
 Sum(int&amp; v): value(v) {}
 void operator()(int data) const { value += data;}

 int&amp; value;
};

int main()
{
 int array[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };

 int total = 0;
 std::for_each( array, array+10, Sum(total));

 std::cout &lt;&lt; total &lt;&lt; std::endl;
}
</code></pre>

blocks|key|1247692|text|每次都会调用myFunction，而total是函数的本地函数...请尝试将total标记为static：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1247693|static+int+total+=+0|code-block|syntax|javascript|1247694|entityMap^0|6|A|I|5|12|5|1A|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

<code>myFunction</code> is being called each time and <code>total</code> is local to the function... try marking <code>total</code> as <code>static</code> instead:

<pre><code>static int total = 0
</code></pre>

blocks|key|1250071|text|当您声明一个变量(即int+total)时，它在其作用域的持续时间内一直存在(通常相当于周围最接近的一对{和}。因此，在您的函数myFunction中，当函数返回时，total将不复存在。它每次调用返回一次--也就是数组中的每个元素返回一次。为了实际求和变量的值(或者将变量保留在myFunction末尾之外)，必须给它一个更大的作用域。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1250072|有两种相关的方法可以做到这一点。一种是“好”的方式，另一种是“简单但风格不好”的方式。好的方法是使用函数器或上下文对象--@Martin已经发布了一个示例。“坏”的方法是将int+total标记为static。如果你的代码是单线程的，它会在你第一次使用它的时候工作。然后就再也不会了。如果有人建议..。，不要这样做。:)|BOLD|1250073|entityMap^0|A|9|1G|1|1I|1|1S|A|2B|5|3W|A|0|2E|9|2Q|6|48|8|0^^$0|@$1|2|3|4|5|6|7|K|8|@$9|L|A|M|B|C]|$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|H]]|D|@]|E|$]]|$1|I|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|J|$]]

When you declare a variable (i.e. <code>int total</code>) it exists for the duration of its scope (usually equivalent to the nearest surrounding pair of <code>{</code> and <code>}</code>. So, in your function <code>myFunction</code>, <code>total</code> ceases to exist when the function returns. It returns once per call--once per element in your array, that is. In order to actually sum its values (or otherwise preserve a variable beyond the end of <code>myFunction</code>, you must give it a broader scope.

There are two relevant ways to do this. One is a "good" way, and one is an "easier-but-badly-styled" way. The good way involves a functor or context object--@Martin has already posted an example. The "bad" way is marking <code>int total</code> as <code>static</code>. It'll work the first time you use it, if your code is single-threaded... and then never again. If somebody suggests it... don't do it. :)

blocks|key|4873340|text|您需要在函数调用之间使总持久化。除了将中间结果添加到中间结果之外，还需要单独访问结果，这将完全排除(至少直接使用)函数--您实际上需要一个类来代替。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4873341|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|$]]

You need to make the total persistent across function calls. You also need to access the result separately from adding the intermediate results to it, which rules out (at least straightforward use of) a function at all -- you really need a class instead.

blocks|key|4873392|text|每次调用该函数时，您都会将'total‘的值重置为0。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4873393|将其声明为'static'|offset|length|style|CODE|4873394|void+myFunction+(int+i)+{
++++static+int+total+=+0;
++++total+%2B=+i;
++++cout+<<+total;
}|code-block|syntax|javascript|4873395|编辑：|BOLD|4873396|或者，如果您希望稍后访问“total”的值，则需要使用全局变量(某种类型！可以在类或函数器中！不要激怒我！)，或者只使用for循环，并将其作为指针传递(即，不使用for_each)：|4873397|void+myFunction+(int+i,+int+*+p_total)+{
++++//No+initialization
++++*p_total+%2B=+i;
++++cout+<<+*p_total;
}


int+main()+{
++++int+array[]+=+{+1,+2,+3,+4,+5,+6,+7,+8,+9,+10+};
++++int+total+=+0;

++++for(int+i+=+0;+i+<+10,+i%2B%2B)
+++++++myFunction(array[i],+&total);

++++//total+is+now+55

++++return+0;
}|4873398|注意:我是一名试图学习C%2B%2B的C程序员。这就是我的做法(非常类似于C语言)，这可能不是标准的C%2B%2B方式。|4873399|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cprogramming.com/tutorial/statickeyword.html^0|0|5|8|5|8|0|0|0|0|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|14|8|@$D|15|E|16|F|G]]|9|@$D|17|E|18|1|19]]|A|$]]|$1|H|3|I|5|J|7|1A|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1B|8|@$D|1C|E|1D|F|O]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1E|8|@]|9|@]|A|$]]|$1|R|3|S|5|J|7|1F|8|@]|9|@]|A|$K|L]]|$1|T|3|U|5|6|7|1G|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|W|$X|$5|Y|Z|10|A|$11|12]]]]

You reset the value of 'total' to 0 each time you call the function. 

Declare it <a href="http://www.cprogramming.com/tutorial/statickeyword.html" rel="nofollow"><code>'static'</code></a> 

<pre><code>void myFunction (int i) {
 static int total = 0;
 total += i;
 cout &lt;&lt; total;
}
</code></pre>

EDIT:

Alternatively, if you want to access the value of 'total' later, you will need to either use a global variable (of some kind! Could be in a class or functor! don't flame me!), or just use a for loop, and pass it in as a pointer (i.e., not use for_each):

<pre><code>void myFunction (int i, int * p_total) {
 //No initialization
 *p_total += i;
 cout &lt;&lt; *p_total;
}


int main() {
 int array[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
 int total = 0;

 for(int i = 0; i &lt; 10, i++)
 myFunction(array[i], &amp;total);

 //total is now 55

 return 0;
}
</code></pre>

Note: I'm a C programmer trying to learn C++. This is how I would do it (which is very C-like), it might not be the standard C++ way.

blocks|key|1247790|text|total是一个具有自动存储持续时间的变量。每次调用myFunction()时，都会创建一个新的total并将其初始化为0。您可以：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1247791|1247792|为total提供静态存储持续时间(使用static关键字)，但是您不能将其值赋给任何东西，因为它仍然是局部作用域。如果你想重用这个函数，无论如何都不是一个好主意。|unordered-list-item|1247793|使total成为一个全局变量。如果你想重用这个function|1247794|make作为一个“函数器”，这也是一个坏主意，就像Martin+York的回答中所描述的那样。这是可重用程度最高的implementation|1247795|1247796|但是，我选择的解决方案是“你问错了问题”，你应该使用std::accumulate()|1247797|#include+<iostream>
#include+<numeric>

int+main()+{
++int+array[]+=+{+1,+2,+3,+4,+5,+6,+7,+8,+9,+10+};

++int+total+=+std::accumulate(array,+array%2B10,+0);
++std::cout+<<+total+<<+'\n';
++return+0;
}|code-block|syntax|javascript|1247798|entityMap^0|0|5|Q|C|1C|5|0|0|1|5|J|6|0|1|5|0|0|0|Q|H|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|14|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|J|3|K|5|I|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|L|3|M|5|I|7|1D|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|S|7|1I|8|@]|D|@]|E|$T|U]]|$1|V|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|W|$]]

<code>total</code> is a variable with automatic storage duration. Every time <code>myFunction()</code> is called, a new <code>total</code> is created and initialized to 0. You could:

<ul>
<li>give <code>total</code> static storage duration (with the <code>static</code> keyword), but you won't be able to assign its value to anything, because it is still local scope. A bad idea if you want to reuse this function, anyhow.</li>
<li>make <code>total</code> a global variable. Also a bad idea if you want to reuse this function</li>
<li>make a "functor", as described in Martin York's answer. This is the most reusable implementation</li>
</ul>

But, my chosen solution is "you're asking the wrong question" and you should be using <code>std::accumulate()</code>:

<pre><code>#include &lt;iostream&gt;
#include &lt;numeric&gt;

int main() {
 int array[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };

 int total = std::accumulate(array, array+10, 0);
 std::cout &lt;&lt; total &lt;&lt; '\n';
 return 0;
}
</code></pre>

I am try to make the myFunction give me a sum of the values in the array, but I know I can not use a return value, and when I run my program with the code as so all I get is a print out of the values and no sum why is that?

<pre><code>void myFunction (int i) {
int total = 0;
total += i;
cout &lt;&lt; total;
}


int main() {
int array[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };

for_each( array, array+10, myFunction);


return 0;
}
</code></pre>

C++ for_each() and the function pointer

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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我试图让myFunction给我数组中的值的和，但我知道我不能使用返回值，当我用代码运行我的程序时，我得到的只是打印出的值而没有和，这是为什么？void myFunction (int i) {int total = 0;total += i;cout << total;}int main() {int array[]...

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