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In the documentation it states “the derived function ‘[f;ff] has the rank of ff and returns f ff[x;y;z]“
If we use the following examples:
q)ff:{[w;x;y;z]w+x+y+z}
q)ff[1;2;3;4]
10
If the syntax returns f ff[x;y;z] then essentially the result for ff[w;x;y;z] is our x parameter for f.
We know that ff[1;2;3;4] = 10
q)f:{2*x}
q)f[10]
20
We use composition to combine these steps together:
q)'[f;ff][1;2;3;4]
20
If you have any questions about this, please let me know!
Thanks,
Megan