Re: The += operator, again

[Domingo Alvarez Duarte <mingodad@...>]

Hello Sam !

First of all thanks for commenting !

I didn't tried something like this before so I wrote this small program and tried to compile/run:

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var val = 3;
var tab = [1,2];
tab[++#tab] = val;

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Output:

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ljs: test-pp.ljs:3: unexpected symbol near '#'

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But it works this way showing a bug ?:

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var val = 3;
var tab = [1,2];
tab[++(#tab)] = val;

====

Output:

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ljs test-pp.ljs
1    1
2    3

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The way it's implemented right now reject it, but a way I usually use it in situations like this is as follow:

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var tab = {"one": 1, "two" : 2, "three": 3, "four": 4};
var ary = [];
var idx = 0;
for(k,v in pairs(tab)) ary[++idx] = [k,v];
table.sort(ary, function(a,b){ return a[2] < b[2];});

for( idx, v in ipairs(ary)) print(idx, v[1], v[2]);

====

Output:

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ljs test-pp2.ljs
1    one    1
2    two    2
3    three    3
4    four    4

===

I appreciate your comments.

Again thanks in advance for your time and attention !

On 05/12/2018 21:16, Sam Putman wrote:

On Wed, Dec 5, 2018 at 2:00 AM Domingo Alvarez Duarte <mingodad@gmail.com> wrote:

Hello !

I've updated the repositories with fixes for compound operators on upvalues and also added pre/pos increment operators "++/--".

If you don't mind my asking, what does this do?

tab[++#tab] = val

let's say #tab was 3, what happens?