• FeepingCreature (39/39) Jan 24 2019 A problem that's come up before is how to specify builders in a
• Eugene Wissner (7/46) Jan 25 2019 $ is a simple function in Haskell that is defined in a library,
• Kagamin (7/15) Jan 25 2019 This is shorter and cleaner:
• bauss (3/22) Jan 25 2019 And easy to implement:
• FeepingCreature (12/21) Jan 25 2019 Yeah but it doesn't look like assignment. And it's only as

FeepingCreature <feepingcreature gmail.com> writes:

A problem that's come up before is how to specify builders in a 
typechecked manner. In my language syntax experiment, Neat, I'd 
copied Haskell's '$' operator, which means "capture everything to 
the right of here in one set of parens", in order to avoid 
writing parentheses expressions:

     foo(2 + 2);
     // becomes
     foo $ 2 + 2;

But what if we wanted the *opposite*? What if we wanted to 
capture everything to the *left* of an operator in one set of 
parens? Enter the tail operator:

     (2 + 2).foo;
     // becomes


Why is this useful?

Well, we've had the debate before about struct initializers. The 
irksome thing is that the language is *almost* there! We can 
already write function calls as assignments:

     foo(5);
     // becomes
     foo = 5;

The main problem is that there's no way in the language to chain 
assignments together in that syntax. If we could, we could 
generate very clean looking builder expressions. But

     (((Foo.build
       .a = 3)
       .b = 4)
       .c = 5)
       .value;

Just doesn't look good and isn't as easy to modify as it should 
be either.

But with the tail operator, this becomes:

     Foo.build





Which is pure, typechecked, easy to extend and still readable.

What do you think?

Eugene Wissner <belka caraus.de> writes:

On Friday, 25 January 2019 at 07:24:10 UTC, FeepingCreature wrote:
 A problem that's come up before is how to specify builders in a 
 typechecked manner. In my language syntax experiment, Neat, I'd 
 copied Haskell's '$' operator, which means "capture everything 
 to the right of here in one set of parens", in order to avoid 
 writing parentheses expressions:

     foo(2 + 2);
     // becomes
     foo $ 2 + 2;

 But what if we wanted the *opposite*? What if we wanted to 
 capture everything to the *left* of an operator in one set of 
 parens? Enter the tail operator:

     (2 + 2).foo;
     // becomes


 Why is this useful?

 Well, we've had the debate before about struct initializers. 
 The irksome thing is that the language is *almost* there! We 
 can already write function calls as assignments:

     foo(5);
     // becomes
     foo = 5;

 The main problem is that there's no way in the language to 
 chain assignments together in that syntax. If we could, we 
 could generate very clean looking builder expressions. But

     (((Foo.build
       .a = 3)
       .b = 4)
       .c = 5)
       .value;

 Just doesn't look good and isn't as easy to modify as it should 
 be either.

 But with the tail operator, this becomes:

     Foo.build





 Which is pure, typechecked, easy to extend and still readable.

 What do you think?

$ is a simple function in Haskell that is defined in a library, 
the same as '+', '-' are just functions and not special operators 
as in D. In D you have to define each operator in the language 
itself and the compiler should understand them. Defining a 
language construct for each use case doesn't seems pretty to me, 
it makes the grammar overcomplicated.

Kagamin <spam here.lot> writes:

On Friday, 25 January 2019 at 07:24:10 UTC, FeepingCreature wrote:
 But with the tail operator, this becomes:

     Foo.build





 Which is pure, typechecked, easy to extend and still readable.

 What do you think?

This is shorter and cleaner:

     Foo.build
       .a(3)
       .b(4)
       .c(5)
       .value;

bauss <jj_1337 live.dk> writes:

On Friday, 25 January 2019 at 08:33:51 UTC, Kagamin wrote:
 On Friday, 25 January 2019 at 07:24:10 UTC, FeepingCreature 
 wrote:
 But with the tail operator, this becomes:

     Foo.build





 Which is pure, typechecked, easy to extend and still readable.

 What do you think?

 This is shorter and cleaner:

     Foo.build
       .a(3)
       .b(4)
       .c(5)
       .value;

And easy to implement:

https://run.dlang.io/is/0Z0GKb

FeepingCreature <feepingcreature gmail.com> writes:

On Friday, 25 January 2019 at 10:31:15 UTC, bauss wrote:
 This is shorter and cleaner:

     Foo.build
       .a(3)
       .b(4)
       .c(5)
       .value;

 And easy to implement:

 https://run.dlang.io/is/0Z0GKb

Yeah but it doesn't look like assignment. And it's only as 
readable because the examples are oversimplified.

     Foo.build
       .property(source.property.call(bla, bla, 
bla).map!"a.BlaAttribute"))

And it becomes much harder to understand the builder principle as 
compared to

     Foo.build

bla).map!"a.BlaAttribute"

We have operators for a reason.