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Exclude backreferences

importance: 5

In simple cases of circular references, we can exclude an offending property from serialization by its name.

But sometimes there are many backreferences. And names may be used both in circular references and normal properties.

Write replacer function to stringify everything, but remove properties that reference meetup:

let room = {
  number: 23
};

let meetup = {
  title: "Conference",
  occupiedBy: [{name: "John"}, {name: "Alice"}],
  place: room
};

// circular references
room.occupiedBy = meetup;
meetup.self = meetup;

alert( JSON.stringify(meetup, function replacer(key, value) {
  /* your code */
}));

/* result should be:
{
  "title":"Conference",
  "occupiedBy":[{"name":"John"},{"name":"Alice"}],
  "place":{"number":23}
}
*/
let room = {
  number: 23
};

let meetup = {
  title: "Conference",
  occupiedBy: [{name: "John"}, {name: "Alice"}],
  place: room
};

room.occupiedBy = meetup;
meetup.self = meetup;

alert( JSON.stringify(meetup, function replacer(key, value) {
  return (key != "" && value == meetup) ? undefined : value;
}));

/*
{
  "title":"Conference",
  "occupiedBy":[{"name":"John"},{"name":"Alice"}],
  "place":{"number":23}
}
*/

Here we also need to test key=="" to exclude the first call where it is normal that value is meetup.