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Home/ Questions/Q 7603381
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T23:38:30+00:00 2026-05-30T23:38:30+00:00

When I try and compile this code (VS2010) I am getting the following error:

  • 0

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); }));
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T23:38:31+00:00Added an answer on May 30, 2026 at 11:38 pm

    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.

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