DIP 1027---String Interpolation---Community Review Round 1

[Jacob Carlborg]

On 2019-12-17 11:14, Patrick Schluter wrote:

> Yes, and that was the point of the exercice. Transforming the 
> interpolated "string" into a string evaluates the values of the 
> variables at the moment of that transformation. From then on, there is 
> no interpolation possible as it a simple string.
> I suppose that what aliak leant in his comments was that the string then 
> could still be used for interpolation.
> 
> hypothetic scenario of what I interpreted what aliak wanted (and it 
> won't compile, it's just for illustration)
> 
> int apple;
> ...
>    string my_is = i"apple=${apple}";
> 
> ...
> 
>    apple = 36;
>    writeln(my_is);
> 
> would print "apple=36"
> 
> with this DIP, when you do this
> 
> int apple;
> ...
>    string my_is = i"apple=${apple}".format;
> 
> ...
> 
>    apple = 36;
>    writeln(my_is);
> 
> will print "apple=0" because my_is will contain the string "apple=0"

I don't see how lowering to a tuple will change anything. It's not 
possible to store expressions. If you have the expression `3 + 4` than 
that will be evaluated and `7` is what's left. Unless the expression is 
wrapped in a lambda.

> The basic thing here is what Walter said above:
> 
> interpolated strings are not string literals, they can't be. They are code.

Of course they can. They can be whatever we decide them to be.

-- 
/Jacob Carlborg