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Re: Every expression has a type?

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  Re: Every expression has a type? Wayne <wayne@nospam.invalid> - 2017-10-31 03:07 -0400
    Re: Every expression has a type? Eric Sosman <esosman@comcast-dot-net.invalid> - 2017-10-31 12:00 -0400
    Re: Every expression has a type? Eric Sosman <esosman@comcast-dot-net.invalid> - 2017-10-31 13:06 -0400

#37616 — Re: Every expression has a type?

On 10/30/2017 8:29 AM, Stefan Ram wrote:
>   I was flabbergasted. I had made some handwritten notes
>   during my course, and one read:
> 
> _Jeder_ Ausdruck hat einen Typ.
> 
>   indicating that I wrote "Jeder Ausdruck hat einen Typ."
>   ("Every expression has a type.") in my published HTML
>   course notes, and that there was something wrong with
>   "Jeder" ("Every").
> 
>   But I can't remember what I was thinking of when I wrote
>   this. How can there be an expression in Java without a type?
>   So, eventually, I discarded my note.
> 
>   Still, I was wondering ...
> 
>   Then it struck my mind!
> 
>   The JLS says:
> 
>       THE Java programming language is a statically typed
>       language, which means that every variable and every
>       expression has a type that is known at compile time.
> 
>   But what's the type of a call of a method with the result
>   (jls9:8.4.5) »void«?
> 
>   In Java, »void« is not a type!
> 
>   So, »java.lang.System.out.println()« is an expression,
>   but does it have a type?
> 

See 15.1, where all that is spelled out (Java 9 version):
<https://docs.oracle.com/javase/specs/jls/se9/html/jls-15.html#jls-15.1>

"An expression denotes nothing if and only if it is a method invocation (§15.12) that
invokes a method that does not return a value, that is, a method declared void
(§8.4). Such an expression can be used only as an expression statement (§14.8) or as
the single expression of a lambda body (§15.27.2), because every other context in
which an expression can appear requires the expression to denote something. ..."

So, an expression consisting of a void method invocation has no result and thus
no type.

-- 
Wayne

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#37628

On 10/31/2017 11:03 AM, Stefan Ram wrote:
> Wayne <wayne@nospam.invalid> writes:
>> See 15.1, where all that is spelled out (Java 9 version):
> ...
>> (§8.4). Such an expression can be used only as an expression statement (§14.8) or as
>> the single expression of a lambda body (§15.27.2), because every other context in
>> which an expression can appear requires the expression to denote something. ..."
> 
>    My compiler is fine with this one:
> 
> public final class Main
> {
> 
>    private static void nothing() {}
> 
>    public static void main( final java.lang.String[] args )
>    { for( nothing(); true; nothing() )break; }}

     JLS 14.14.1 tells us that the first and third pieces of
"BasicForStatemet" are "ForInit" and "ForUpdate", respectively.
Both of these are "StatementExpressionList"s, which are comma-
separated lists of "StatementExpression"s.  A "StatementExpression"
is that part of an "ExpressionStatement" preceding the semicolon
(14.8), so all is well.

     (By the way, I just found yet another infinity of angels holding
a cotillion on that zero-sized pinhead.)

-- 
esosman@comcast-dot-net.invalid
Eleven hundred seventy-seven days to go.

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#37630

On 10/31/2017 11:13 AM, Stefan Ram wrote:
> [...]
>    We can have types of expressions, which do not have any
>    value at all. For example consider this program:
> 
> public final class Main
> {
> 
>    public static double test( final int i ){ return i; }
> 
>    public static void main( final java.lang.String[] args ) {} }
> 
>    The compiler will check that the type »int« of the return
>    expression »i« is a subtype of the type of the return type
>    »double«. So, »i« has a type. But »i« does never have a value
>    in this program.

     The compiler doesn't require that the type of `i' be a subtype
of the return type.  If it did, the example would not compile: `int'
is not a subtype of `double'.  The compiler requires, rather, that
the expression in the `return' produce a value that is convertible
to the method's returned type, and that's a far weaker requirement.

>    So, »i« is an expression, it has a type, but not a value.

     I think all you're saying is that code not executed doesn't
execute.  That is, what you say of `i' is also true of `args':
If the main() method is not called, `args' has no value.

>    So, it actually would be perfectly fine for the call of
>    a void-method to have a type but not a value, as is the
>    case in many other languages like, for example, C, where
>    »void« is a type.

     `void' is a type in C, but not a type in Java.  `if' can test an
`int' expression in C, but not in Java.  `->' is an operator in C, but
not in Java.  `goto' is a statement in C, but a reserved and meaningless
word in Java.  So what?  Different languages assign different meanings
to similar symbols.  `Gift' in English is something that you'd like to
receive, but that in German you'd probably refuse.[*]

     [*] https://translate.google.com/#de/en/Gift

-- 
esosman@comcast-dot-net.invalid
Eleven hundred seventy-seven days to go.

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