json::parse(...) vs json j; j.parse(...)

[oneill1979]

I was wondering if there is a reason that a call to parse on a json object just returns a null json object? The same input seems to parse without issue using the jason::parse call.

For example the following program:

{
    std::string input("{\"happy\":true,\"pi\":3.141}");
    json j;
    j.parse(input);
    std::cout << j.dump() << std::endl;
    j=json::parse(input);
    std::cout << j.dump() << std::endl;
}

Results in (osx):

null
{"happy":true,"pi":3.141}

[nlohmann]

In the upper example, you create a json object j which is initially null. In line 4, you basically call the static function json::parse(input); which returns the result of the deserialization. As this result is not stored, j remains unchanged. In line 6, you make the same call, but store the result in j, yielding the expected result.

Again: parse is not a member function, but a static function. Calling it as a member function is syntactically OK, but does not change the object you call it from.

[gregmarr]

The parse function is static. There is no such thing as calling it "on an object", except that C++ allows you to identify the static function to call using an object instead of the class name.
Your code is equivalent to this:

std::string input("{\"happy\":true,\"pi\":3.141}");
json::parse(input);
std::cout << json().dump() << std::endl;
json j=json::parse(input);
std::cout << j.dump() << std::endl;

[gregmarr]

Heh, I saw @nlohmann's answer appear just as I clicked the Comment button. :)

[nlohmann]

Same here, but Github claims I was 3 seconds faster ;)

[oneill1979]

Ah, thank you. I had not noticed that it was a static method.