Question

so for instance

f() -> 3.
x() -> F = fun() -> f() end,
       A = f(),
       B = ?MODULE:f(),
       C = F().

right after F hash been defined, a new version of code for f() is defined as follows f()-> three and loaded. What would be the value of C.

Answer 1

if you cant read the whole answer Update context: Answer is "three"

EDIT: Sorry The answer is misleading, here is the right answer: As the F is just defined but the function is not evaluated, since we are making a fully qualified call before evaluating F; the new code will be loaded and the value will be "three"

So, i posted the question in erlang mailing list, i got the reply with this answer

Funs are always bound to the code they are initially loaded from. Only name look-ups are affected by code loading AFAICT. This has been a source of exceptions in our previous development, because the first time you load new code, F is still valid, but the second time you load new code, the old code is purged and F is now invalid. Any call to it will generate an exception. We ended up wrapping our needs for lambdas in a module with state instead. Not the most elegant, but allows us to get "dynamic lambdas." If all you need in the fun is module:function, you can use a tuple for that instead of a fun. Sincerely, jw

Answer 2

例如,B的价值只有三倍。模块的最新版本只有在向单元发出外部电话时才能完成。这使热密码升级原子操作,你要么重新执行旧的或新的法典,从来都不执行。

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