On 2021-12-01 23:36, Chris Angelico wrote:
That's exactly why it's such a close parallel. The late-evaluated default is just code, nothing else. It's not "stored" in any way - it is evaluated as part of the function beginning execution.
But it IS stored! There is no way for it to be evaluated without it being stored! I know we're talking past each other here but it is quite obvious that something has to be stored if it is going to be evaluated later. You can say that it is "just code" but that doesn't change the fact that that code has to be stored. You can say that it is just prepended to the function body but that's still storing it. That is still not parallel to a ternary operator in which no part of the expression is EVER re-executed unless control flow causes execution to return to that same source code line and re-execute it as a whole. Actually this raises a question that maybe was answered in the earlier thread but if so I forgot: if a function has a late-bound default, will the code to evaluate it be stored as part of the function's code object? -- Brendan Barnwell "Do not follow where the path may lead. Go, instead, where there is no path, and leave a trail." --author unknown