blocks|key|3114709|text|您可以通过在参数列表后使用->+Type来显式指定lambda的返回类型：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3114710|[]()+->+Type+{+}|code-block|syntax|javascript|3114711|但是，如果lambda有一条语句，且该语句是一个返回语句(并且它返回一个表达式)，则编译器可以从该返回表达式的类型推断出返回类型。您的lambda中有多个语句，因此它不会推断类型。|3114712|entityMap^0|D|7|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You can explicitly specify the return type of a lambda by using <code>-&gt; Type</code> after the arguments list:

<pre><code>[]() -&gt; Type { }
</code></pre>

However, if a lambda has one statement and that statement is a return statement (and it returns an expression), the compiler can deduce the return type from the type of that one returned expression. You have multiple statements in your lambda, so it doesn't deduce the type.

blocks|key|233225|text|可以推断lambda的返回类型(在C%2B%2B11中)，但前提是只有一条语句，并且该语句是返回表达式的return语句(例如，初始值设定项列表不是表达式)。如果您有一个多语句lambda，则假定返回类型为void。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|233226|因此，您应该这样做：|233227|++remove_if(rawLines.begin(),+rawLines.end(),+[&expression,+&start,+&end,+&what,+&flags](const+string&+line)+->+bool
++{
++++start+=+line.begin();
++++end+=+line.end();
++++bool+temp+=+boost::regex_search(start,+end,+what,+expression,+flags);
++++return+temp;
++})|code-block|syntax|javascript|233228|但实际上，您的第二个表达式的可读性要好得多。|233229|entityMap^0|1C|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

The return type of a lambda (in C++11) can be deduced, but only when there is exactly one statement, and that statement is a <code>return</code> statement that returns an expression (an initializer list is not an expression, for example). If you have a multi-statement lambda, then the return type is assumed to be void.

Therefore, you should do this:

<pre><code> remove_if(rawLines.begin(), rawLines.end(), [&amp;expression, &amp;start, &amp;end, &amp;what, &amp;flags](const string&amp; line) -&gt; bool
 {
 start = line.begin();
 end = line.end();
 bool temp = boost::regex_search(start, end, what, expression, flags);
 return temp;
 })
</code></pre>

But really, your second expression is a lot more readable.

blocks|key|2900527|text|当仍然返回时，您可以有多个语句：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2900528|[]()+->+your_type+{return+(
++++++++your_statement,
++++++++even_more_statement+=+just_add_comma,
++++++++return_value);}|code-block|syntax|javascript|2900529|http://www.cplusplus.com/doc/tutorial/operators/#comma|offset|length|2900530|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|1I|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@]|9|@$I|U|J|V|1|W]]|A|$]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|H]]]]

You can have more than one statement when still return:

<pre><code>[]() -&gt; your_type {return (
 your_statement,
 even_more_statement = just_add_comma,
 return_value);}
</code></pre>

<a href="http://www.cplusplus.com/doc/tutorial/operators/#comma" rel="noreferrer">http://www.cplusplus.com/doc/tutorial/operators/#comma</a>

When I try and compile this code (VS2010) I am getting the following error:
<code>error C3499: a lambda that has been specified to have a void return type cannot return a value</code>

<pre><code>void DataFile::removeComments()
{
 string::const_iterator start, end;
 boost::regex expression("^\\s?#");
 boost::match_results&lt;std::string::const_iterator&gt; what;
 boost::match_flag_type flags = boost::match_default;
 // Look for lines that either start with a hash (#)
 // or have nothing but white-space preceeding the hash symbol
 remove_if(rawLines.begin(), rawLines.end(), [&amp;expression, &amp;start, &amp;end, &amp;what, &amp;flags](const string&amp; line)
 {
 start = line.begin();
 end = line.end();
 bool temp = boost::regex_search(start, end, what, expression, flags);
 return temp;
 });
}
</code></pre>

How did I specify that the lambda has a 'void' return type. More-over, how do I specify that the lambda has 'bool' return type?

UPDATE

The following compiles. Can someone please tell me why that compiles and the other does not?

<pre><code>void DataFile::removeComments()
{
 boost::regex expression("^(\\s+)?#");
 boost::match_results&lt;std::string::const_iterator&gt; what;
 boost::match_flag_type flags = boost::match_default;
 // Look for lines that either start with a hash (#)
 // or have nothing but white-space preceeding the hash symbol
 rawLines.erase(remove_if(rawLines.begin(), rawLines.end(), [&amp;expression, &amp;what, &amp;flags](const string&amp; line)
 { return boost::regex_search(line.begin(), line.end(), what, expression, flags); }));
}
</code></pre>

Explicit Return Type of Lambda

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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当我尝试编译这段代码(VS2010)时，我得到了以下错误：error C3499: a lambda that has been specified to have a void return type cannot return a valuevoid DataFile::removeComments(){  str...

问Lambda的显式返回类型
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