I have two canisters, one calls another. Both have enough free memory (both main one and stable one), and do some computations that normally don’t lead to exceptional conditions (e.g. they don’t divide by zero). Can an intercanister call fail? If yes, what may be the reasons of the failure?

Are intercanister calls safe?

free June 10, 2024, 10:29am 5

Work is in progress. See this thread.

It’s probably still a couple of months out.

2 Likes