Breaking ";" rule with lambda functions

[pascal111]

On Monday, 1 August 2022 at 15:08:04 UTC, ag0aep6g wrote:
> On Monday, 1 August 2022 at 14:52:03 UTC, pascal111 wrote:
>> If `foo => bar` == `(foo) { return bar; }`, then  `foo => bar` 
>> is a function. "=>" is not an operator, it's a special symbol 
>> for lambda "function".
>>
>> If A == B, so A's types is the same of B's type. How can it be 
>> withstanding `foo => bar` == `foo => bar` == `(foo) { return 
>> bar; }` and `foo => bar` is an expression and the other is a 
>> function?!! no sense.
>
> `foo => bar` and `(foo) { return bar; }` are both function 
> literals, and they're both expressions. The concepts are not 
> mutually exclusive.

Surely, because it seems that you are real man, your word must be 
taken. Isn't `(foo) { return bar; }` an anonymous function or am 
I a wrong?!! It IS a function, not an expression.