Optional output file values

[skrowten-hermit]

I have been trying various options with this handy library. But I hit a roadblock with a certain scenario. I would like to ask the program to generate output files using the following:

options.add_options()
    ("s,source", "Source input file", 
                 cxxopts::value<std::string>())
    //("d,debug", "Enable debugging") // a bool parameter
    ("p,splitoutput", "Split the output to separate files", 
                      cxxopts::value<bool>()->default_value("false"))
    ("o,outfile", "Output file(s), separated by \',\' if multiple output "
                  "files (only if -p or --splitoutput is set) needed. The "
                  "order is - main, submodule1,submodule2, submodule3.",
                  cxxopts::value<std::vector<std::string>>()->default_value("default"))
    ("v,verbose", "Verbose output", 
                  cxxopts::value<bool>()->default_value("false"))
    ("h,help", "Print usage")

Now, this requires me to pass at least one argument after -o or --outfile. I would like to make this optional and assign values internally if the user choses not to. For example,

./program -o
./program -o a.txt,b.txt,c.txt,d.txt
./program -o a.txt,,c.txt,d.txt
./program -o a.txt,,,d.txt

should all be possible.

I did the following (in addition to adding default_value as well):

std::vector<std::string> o_files;
if (result.count("outfile"))
  o_files = result["outfile"].as<std::vector<std::string>>();
else
  o_files.push_back("defaults");

So that I can check for vector with "defaults" and process.

But I get the error:

terminate called after throwing an instance of 'cxxopts::missing_argument_exception'
  what():  Option ‘o’ is missing an argument
Aborted

What is going wrong here? Is default_option meant to work with only bool types? Or am I missing something here?

[jarro2783]

If you want -o to work by itself then you need an implicit argument. A default is so that you still get a value when -o is not provided at all. However it makes parsing a bit difficult because it's ambiguous when you write:

-o argument

It is possible that the user meant -o with implicit value, and argument is a positional argument. So the only way to write that correctly is to use a long option and write --output=argument.