Need public API to free returned strings (e.g. cJSON_Print)

[ewmailing]

There should be a public API to free returned strings like from cJSON_Print().

My usage case is that I'm writing a middleware library that uses cJSON. The user is free to use cJSON elsewhere in their project. I don't necessarily know what memory allocator they are using, so when I call cJSON_Print(), I don't know what free() function to call. cJSON_free is private/internal.

I'm thinking either adding a cJSON_FreeString() which wraps cJSON_free, or making CJSON_free public are the easiest solutions.

Thank you

[DaveGamble]

You should implement cJSON_Hooks and discourage them from doing the same.

[FSMaxB]

I propose adding cJSON_Hooks as optional parameter to all the functions that are doing frees/allocations. That would be part of the breaking changes for 1.0.0

[FSMaxB]

What is passed will be a pointer to cJSON_Hooks and if null, regular malloc and free will be used.

[DaveGamble]

Nice.

[DaveGamble]

While you're there, it might be wise to remove unbuffered printing entirely, especially since you have ensure() working nicely now :)

[ajaybhargav]

I would suggest rename cJSON_Delete to cJSON_DeleteObject and create a new API cJSON_DeleteBuffer which frees the print buffer using hook. Let me know I can create patch for that.

[FSMaxB]

No. By allowing the user to pass in a pointer to an instance of cJSON_Hooks, the user of the library knows which allocator was used for allocating the printbuffer, so no special function is required for that, just use the deallocator that fits the allocator you gave to cJSON for printing.

[mjerris]

Is there currently any way using cJSON_Hooks to pass along another void pointer that could go to the allocator? I can't see a way. This would be necessary to do something like using a memory pool allocator? Obviously this would require major upheaval to the api, but seems if this is going to happen, now would be the time. Is there any interest in looking at that now, or would it be too invasive to make sense?