Support Request: Understanding key() during iteration about rapidyaml HOT 5 CLOSED

ryml::csubstr is a string view, and as such it is NOT zero-terminated. So if you do cout << child.key().str you are printing an unbounded string (or at best bounded at the end of the full source string, if you are lucky). That's why it goes past the end of what you expect: there is no end!

Don't do that.

In your case, just do cout << child.key() (notice it does NOT have .str at the end), which will write only up to the size of the view. Likewise, cout << child.val():

#define RYML_SINGLE_HDR_DEFINE_NOW

#include "ryml_all.hpp"

#include <fstream>
#include <iostream>
#include <sstream>
#include <string>

int main() {
  std::string input{
      R"(
Sinks:
  - Type: "Colorized"
    FormatString: "my format string"
    Filter: "AAA Filter"
  - Type: "File"
    FileName: "my file name"
    Filter: "BBB Filter"
    )"};

  size_t i = 0;
  ryml::Tree tree = ryml::parse_in_arena(ryml::to_csubstr(input));
  for (ryml::ConstNodeRef const& sink : tree["Sinks"]) {
    std::cout << std::endl;
    for (ryml::ConstNodeRef const& child : sink) {
      std::cout << "Sink[" << i << "]: ~~~" << child.key() << "~~~/~~~" << child.val() << "~~~" << std::endl;
    }
    ++i;
  }

  return 0;
}

prints this:

Sink[0]: ~~~Type~~~/~~~Colorized~~~
Sink[0]: ~~~FormatString~~~/~~~my format string~~~
Sink[0]: ~~~Filter~~~/~~~AAA Filter~~~

Sink[1]: ~~~Type~~~/~~~File~~~
Sink[1]: ~~~FileName~~~/~~~my file name~~~
Sink[1]: ~~~Filter~~~/~~~BBB Filter~~~

Or you could do k = child.key(); cout.write(k.str, k.len), which is what ryml's implementation of operator<<(ostream&, ryml::csubstr) does when you call it.

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@floli does this answer your question?

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Thanks! Yes, it adds clarity. Just one more question. What is the best method to get a std::string from a node.val() or node.key() ? Is there a way that does not involve using a new variable? e.g. to be used as a parameter (f(node.key().to_str() when void f(std::string))

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If you want to create a std::string there must always be a new variable, as the tree holds only views. If you want syntactic candy to call your f() function without having to create a named variable, just define an adapter in your code:

std::string tostr(ryml::csubstr s) { return {s.str, s.len}; }

and then just do f(tostr(node.key())).

But note this is a performance antipattern, as the temporary std::string will cause a potential allocation and performance jitter.

I would suggest instead making your f() accept directly the ryml::csubstr, or if you don't want it to be aware of a ryml type, then just make it accept f(const char* str, size_t len) (and again note the str is not zero-terminated).

If you are wondering why ryml does not give you a std::string it's precisely because it is an antipattern forcing allocation for every string, and further there are projects in embedded where std::string just cannot be used. In fact if ryml is rapid it is because in large part it does not use std::string.

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Thanks for your explanations!

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