Description
on C11+ and newer compilers, the =
operator does not behave as expected. The code is listed below
Lines 244 to 261 in ea843a9
This overload is invoked when a JSONVar
is set equal to another JSONVar
by reference like so:
JSONVar a const
char input[] = "{'foo':{'bar': '42'}}";
JSONVar a = JSON.parse(input);
JSONVar b = a["foo"];
Serial.print("b['foo']: ");
Serial.println(b["foo"]);
result:
b['foo']: null
I see a few things wrong with this:
First, swapping the c_json
pointers in the JSONVar
container does nothing as the linked list is all stored in the c_json
structs. The JSONVar
object only is created on the fly when requested via the []
operator, and destroyed when out of scope. See below:
JSONVar JSONVar::operator[](const char* key)
{
Serial.println(key);
if (!cJSON_IsObject(_json)) {
replaceJson(cJSON_CreateObject());
}
cJSON* json = cJSON_GetObjectItemCaseSensitive(_json, key);
if (json == NULL) {
json = cJSON_AddNullToObject(_json, key);
}
return JSONVar(json, _json);
}
Second, the intent of the code in the C11+ rvalue =
overloaded operator is to swap the two objects instead of copy. Why? This is not the same behavior as in older compilers and is not the same behavior if the =
operator is used on a JSONVar
object directly without referencing. This is is incredibly confusing to have an object be copied in one situation, but then be swapped in a different situation. In my opinion the =
operator should always copy, regardless if an lvalue or rvalue is passed.