blocks|key|1033437|text|你不需要手工制作的函数对象，只需要使用std::function，lambda函数是可以转换的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1033438|此示例返回整数标识函数：|1033439|std::function<int+(int)>+retFun()+{
++++return+[](int+x)+{+return+x;+};
}|code-block|syntax|javascript|1033440|entityMap^0|J|D|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You don't need a handcrafted function object, just use <code>std::function</code>, to which lambda functions are convertible:

This example returns the integer identity function:

<pre><code>std::function&lt;int (int)&gt; retFun() {
 return [](int x) { return x; };
}
</code></pre>

blocks|key|1033508|text|您可以从其他lambda函数返回lambda函数，因为您不应该显式指定lambda函数的返回类型。只需在全局范围内编写类似的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1033509|+auto+retFun+=+[]()+{
+++++return+[](int+x)+{return+x;};
+};|code-block|syntax|javascript|1033510|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 can return lambda function from other lambda function, since you should not explicitly specify return type of lambda function. Just write something like that in global scope:

<pre><code> auto retFun = []() {
 return [](int x) {return x;};
 };
</code></pre>

blocks|key|818493|text|虽然这个问题特别问到了C%2B%2B11，但为了让其他偶然发现这一点的人能够使用C%2B%2B14编译器，C%2B%2B14现在允许为普通函数推断返回类型。因此，问题中的示例只需通过删除->+decltype就可以根据需要进行调整。函数参数列表后面的子句：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|818494|auto+retFun()
{
++++return+[](int+x)+{+return+x;+}
}|code-block|syntax|javascript|818495|但是，请注意，如果函数中出现多个return+<lambda>;，这将不起作用。这是因为对返回类型推导的限制是，所有返回语句必须返回相同类型的表达式，但是编译器为每个lambda对象指定了自己的唯一类型，因此return+<lambda>;表达式将具有不同的类型。|818496|entityMap^0|29|B|0|0|G|G|2W|G|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

Though the question specifically asks about C++11, for the sake of others who stumble upon this and have access to a C++14 compiler, C++14 now allows deduced return types for ordinary functions. So the example in the question can be adjusted just to work as desired simply by dropping the <code>-&gt; decltype</code>... clause after the function parameter list:

<pre><code>auto retFun()
{
 return [](int x) { return x; }
}
</code></pre>

Note, however, that this will not work if more than one <code>return &lt;lambda&gt;;</code> appears in the function. This is because a restriction on return type deduction is that all return statements must return expressions of the same type, but every lambda object is given its own unique type by the compiler, so the <code>return &lt;lambda&gt;;</code> expressions will each have a different type.

blocks|key|818506|text|你应该这样写：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|818507|auto+returnFunction+=+[](int+x){
++++return+[&x](){
++++++++return+x;
++++}();
};|code-block|syntax|javascript|818508|要将您的返回值作为一个函数来获取，并像这样使用它：|818509|int+val+=+returnFunction(someNumber);|818510|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You should write like this:

<pre class="lang-cpp prettyprint-override"><code>auto returnFunction = [](int x){
 return [&amp;x](){
 return x;
 }();
};
</code></pre>

to get your return as a function, and use it like:

<pre class="lang-cpp prettyprint-override"><code>int val = returnFunction(someNumber);
</code></pre>

blocks|key|1033590|text|例如，如果您没有c%2B%2B+11，并且正在微控制器上运行您的c%2B%2B代码。您可以返回一个空指针，然后执行强制转换。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1033591|void*+functionThatReturnsLambda()
{
++++void(*someMethod)();

++++//+your+lambda
++++someMethod+=+[]()+{

++++++++//+code+of+lambda

++++};

++++return+someMethod;
}


int+main(int+argc,+char*+argv[])
{

++++void*+myLambdaRaw+=+functionThatReturnsLambda();

++++//+cast+it
++++auto+myLambda+=+(void(*)())myLambdaRaw;

++++//+execute+lambda
++++myLambda();
}|code-block|syntax|javascript|1033592|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|$]]

If you do not have c++ 11 and are running your c++ code on micro controllers for example. You can return a void pointer and then perform a cast.
<pre><code>void* functionThatReturnsLambda()
{
 void(*someMethod)();

 // your lambda
 someMethod = []() {

 // code of lambda

 };

 return someMethod;
}


int main(int argc, char* argv[])
{

 void* myLambdaRaw = functionThatReturnsLambda();

 // cast it
 auto myLambda = (void(*)())myLambdaRaw;

 // execute lambda
 myLambda();
}
</code></pre>

I wonder if it's possible to write a function that returns a lambda function in C++11. Of course one problem is how to declare such function. Each lambda has a type, but that type is not expressible in C++. I don't think this would work:

<pre><code>auto retFun() -&gt; decltype ([](int x) -&gt; int)
{
 return [](int x) { return x; }
}
</code></pre>

Nor this:

<pre><code>int(int) retFun();
</code></pre>

I'm not aware of any automatic conversions from lambdas to, say, pointers to functions, or some such. Is the only solution handcrafting a function object and returning it?

Function returning a lambda expression

微控制器

编译器

我想知道是否有可能用C++11编写一个返回lambda函数的函数，当然其中一个问题是如何声明这样的函数。每个lambda都有一个类型，但是该类型在C++中是不可表达的。我不认为这是可行的：auto retFun() -> decltype ([](int x) -> int){    return [](int x) ...

问返回lambda表达式的函数
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