Explicit Return Type of Lambda

问题

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

void DataFile::removeComments()
{
  string::const_iterator start, end;
  boost::regex expression("^\\s?#");
  boost::match_results<std::string::const_iterator> 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(), [&expression, &start, &end, &what, &flags](const string& line)
  {
    start = line.begin();
    end = line.end();
    bool temp = boost::regex_search(start, end, what, expression, flags);
    return temp;
  });
}

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?

void DataFile::removeComments()
{
  boost::regex expression("^(\\s+)?#");
  boost::match_results<std::string::const_iterator> 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(), [&expression, &what, &flags](const string& line)
  { return boost::regex_search(line.begin(), line.end(), what, expression, flags); }));
}

回答1:

You can explicitly specify the return type of a lambda by using -> Type after the arguments list:

[]() -> Type { }

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.

回答2:

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 return 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:

  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;
  })

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

回答3:

You can have more than one statement when still return:

[]() -> your_type {return (
        your_statement,
        even_more_statement = just_add_comma,
        return_value);}

http://www.cplusplus.com/doc/tutorial/operators/#comma

来源：https://stackoverflow.com/questions/9620098/explicit-return-type-of-lambda