Imagine we have a nonempty sequence named \l_tmpa_seq. We want to assign a name \l_tmpb_seq a value which is the result of invoking a function like

\int_eval:n {#1 + 1}

on every member of the sequence.

So take

\seq_set_eq:Nx \l_tmpb_seq  { \l_process_cs:N \l_tmpa_seq }


\cs_new_protected:Npn \l_process_cs:N #1 
    % Assign #1 to \l_name_seq
    % Process the sequence \l_name_seq
    % ...
    % Leave / return the sequence \l_name_seq

and (for brevity; might be a custom function instead)

\cs_generate_variant:Nn \seq_set_eq:NN { Nx }

For now I resort to

  1. mutating a second argument passed to such a function or
  2. invoking \seq_use:Nn \l_name_seq {some_delimiter} as a return statement in an imperative language to serialize the result and invoking \seq_set_split:Nnn \l_another_seq {#1} {some_delimiter} as an assignment statement.

The first approach I would like to avoid completely, whereas the second is cumbersome but admits syntactic constructs like

\cs_return:N \l_name_seq

which could be done in a monadic fashion and

\seq_let:Nx \l_another_seq {\l_someFunction:N #1}

which I will implement to see how it works.

How do you return, serialize, evaluate the result of a function when assigning it to a variable outside that scope?

closed as unclear what you're asking by egreg, Gonzalo Medina, user13907, Sean Allred, Paul Gaborit Jul 4 '15 at 23:31

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  • 2
    Isn't it \seq_set_map:NNn? – egreg Mar 19 '15 at 11:47
  • 1
    Welcome to TeX-sx! At the moment, I'm struggling to understand the question here: I think something like 'How do I arrange \l_tmpb_seq to contain the result of applying some function to each element of \l_tmpa_seq?' but I am far from certain. Could you perhaps add a 'non-programmers' explanation as most of us don't have formal backgrounds in the area! – Joseph Wright Mar 19 '15 at 11:48
  • @egreg I'd forgotten we had that one :-) – Joseph Wright Mar 19 '15 at 11:52
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    BTW, something like \cs_generate_variant:Nn \seq_set_eq:NN { Nx } is definitely wrong as the base argument for x is n, whereas the only variant valid for N is c. – Joseph Wright Mar 19 '15 at 12:16
  • @egreg: Wow, I didn't look through the "Candidates" section of the manual (Sep 15, 2014). This may be exactly what I need. – user74514 Mar 19 '15 at 13:42

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