[英]Typed list typespec never breaks the contract
If you define a typespec and use a different type of parameter it will display an error similar to: 如果定义类型规范并使用其他类型的参数,则将显示类似于以下内容的错误:
binary() ... breaks the contract ... boolean()
For example this typespec: 例如,此typespec:
@spec check?(binary) :: boolean
But it doesnt seem to work for a typed List, or at least, it will never display the warning, if I have a method which receives a list of strings I would define this typespec: 但是它似乎不适用于类型列表,或者至少不会显示警告,如果我有一个接收字符串列表的方法,我将定义此类型规范:
@spec check?([String.t]) :: boolean
I can then define any spec for the list and it will never complain when running a dialyzer, ie: 然后,我可以为列表定义任何规范,并且在运行透析器时也不会抱怨,即:
@spec check?(list(boolean)) :: boolean
@spec check?(list(Conn)) :: boolean
@spec check?(list(number)) :: boolean
@spec check?(list(integer)) :: boolean
Is that intended? 那是故意的吗? it looks like if I defined a list with any type
[any()]
好像我定义了任何类型的列表
[any()]
Is there other way to achieve this? 还有其他方法可以做到这一点吗?
The reason this happens is that all the list types include the empty list as a valid value. 发生这种情况的原因是所有列表类型都将空列表包含为有效值。
For example, in the following case: 例如,在以下情况下:
Dialyzer will conclude that there is a possible solution, namely if the list is empty. Dialyzer将得出一个可能的解决方案,即如果列表为空。 Since Dialyzer only prints warnings if it can conclude that a certain piece of code will always crash, it doesn't print one in this case.
由于Dialyzer仅在可以断定某段代码将始终崩溃的情况下才打印警告,因此在这种情况下不打印任何代码。
I'm not aware of any good solution to this. 我不知道有什么好的解决方案。 If you want to explicitly require non-empty lists, you can use eg
nonempty_list(boolean)
instead of list(boolean)
. 如果要明确要求非空列表,则可以使用
nonempty_list(boolean)
代替list(boolean)
。
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