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Comment by jdougan | Hacker News Reader

Is there a reason you can't do printf("%.*s", strlen, strptr); ?

A non-allocating library would be forced to return to you the unparsed string literal, since returning a parsed string can require an allocation. It might tell you that the literal is valid.

E.g., take the JSON:

  "\"\uD83D\uDCA9\""

There's no (pointer, length) into that that you can then printf(…, ptr, len); you'd get the escapes, raw.

Ofc., there might be situations like debugging where that's fine.

The terminating character is still the closing double quote and not a null, since the library does neither copy out nor alter the input. For example tiny_json replaces the closing quotes to create C strings, but that needs the full file to be in a mutable buffer which can be prohibitive for small controllers reading some config from flash only.

With the "%.*s" format you need no null at the end. It just counts out the characters:

    #include <stdio.h>

    void main()
    {
      char buff[10] = {'R', 'o', 'b', 'o', 't', 't', 'y', 'p', 'e', 's'};
      printf("=>%.*s<=", 4, ((char *)&buff + 3));
    }

prints

    =>otty<=

Ah. Ok. Scanning the length before printing is mandatory then.

You have to do (a variant of) one or the other, no?

@msarnoff had already stipulated that the json lib is returning lengths:

    all the “tokens” are given as byte offsets and lengths

Yes, that’s exactly what I’ve done.

Any other issues with it? Sibling comments mentioned potential unicode issues.

I've only used it in an application where we ensure the data is ASCII-only. The only issue is that I've had to write a bunch of wrapper code around it (for looking up object properties by key, iterating over key/value pairs or array elements, etc.)

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