Breaking ";" rule with lambda functions

Paul Backus snarwin at
Mon Aug 1 19:32:41 UTC 2022

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.

 From [the relevant section of the language spec:][1]

> FunctionLiterals (also known as Lambdas) enable embedding 
> anonymous functions and anonymous delegates directly into 
> expressions. [...]  The type of a function literal is a 
> delegate or a pointer to function.

In other words, a function literal is an expression that 
evaluates to either a delegate or a function pointer.

You are correct that, strictly speaking, it is wrong to say that 
a function literal "is" an anonymous function (rather than 
"refers to" or "points to" one). However, the distinction usually 
does not matter in practice, so most D programmers use the terms 


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