How do I assign the values of a dictionary to the current scope?

Question

Let's say I have a dictionary:

x:(`a`b`c)!(1 2 3)

What are more elegant ways of doing:

a:x`a; b:x`b; c:x`c

I'm looking at both destructuring methods, of the form (a;b;c):value x

and functional methods, of the form : each x (where : refers to the assignment operator).

Answer 1

Is your goal in a local scope only?

   q){[x]
     someCode x;
     (a;b;c);/ Now exist as locals
    } (`a`b`c)!(1 2 3)
   q)a / Global scope not effected.
     'a

It's not something the language caters for.

+ Functions have limits to the number of local variables:

https://code.kx.com/q/basics/function-notation/#variables-and-constants

Running value on a function shows information including a list of locals.

Knowing this in advance likely could help with: (my guesses)

1. Scope .i.e the interpreter knowing the difference between the local x and a possible global x
2. Performance
3. The debugger

https://code.kx.com/q/ref/value/#lambda

q)func:{[x;y] a:1;b:2;c:3}

q)value func
0x0d030902a0030a02a1030b0004
`x`y
`a`b`c
,`
2
3
14 0 13 0 18 0 17 0 22 0 21 5 5
"..func"
""
-1
"{[x;y] a:1;b:2;c:3}"

If you mean in the global scope use set

q)x:(`a`b`c)!(1 2 3)
q)set'[key x;value x]
`a`b`c
q)a
1