Why not all statement are expressions ?

[Stewart Gordon]

On 07/05/2012 18:06, deadalnix wrote:
> Hi,
>
> Working on D I noticed that some statement, notably assert, are expression of type void.
> Why not all statement (that are not expression already) are expression ?

Because it wouldn't make sense.

The semantic of an ExpressionStatement is to evaluate the expression.

Creating a variable in the scope in which an expression appears, and controlling the flow 
of the function in which an expression appears, are beyond the scope of the expression 
evaluation mechanism.

To take an example, what sense would it make if
     return 42
were an expression?  What would its type be?  You would be able to do something like
     auto qwert() {
         return return 105;
     }
to instantly return from the function that called qwert.

OK, so assert, or exception throwing more generally, feels like that kind of thing.  But 
exceptions aren't part of the normal flow of program logic.  Exceptions can be, and indeed 
usually are, thrown in the process of evaluating an expression.  AssertExpression is an 
example of this.

Though this does suggest an answer of "because it can be" for why an assert is an 
expression.  Indeed, you could well ask why an assert is an expression but a throw isn't. 
  Maybe it was thought worth convoluting assert to make for concise code by including it 
in a CommaExpression - but not worth convoluting throw in the same way, since a throw 
throws unconditionally.

Not that it would be much more complicated to make throw an expression.  But you could 
always define a function of your own like

     void throwEx(Throwable t) { throw t; }

Stewart.