CONCATENATED = exists( "BAB" );
if CONCATENATED == 1 then
   SOL = SOL + "-" + "984F8";
   disp( SOL )
   clear BAB;
else
   BAB = 1;
   SOL = "71E17";
end
// Programming Challenge - Concatenation v2 (October 2018)
// A => B; concatenated BAB == result; B,BA,AB ~= result;
// My revised suggestion of a solution to this programming puzzle done in Scilab 6.0.1 code:
// Revision v2: string as argument for function exists and new line inserted at the end of B.

CONCATENATED = exists( "BAB" );
if CONCATENATED == 1 then
   SOL = SOL + "-" + "65E47" + "-" + "AD4D2";
else
   mprintf(' CONCATENATED = exists( ""BAB"" );\n if CONCATENATED == 1 then\n    SOL = SOL + ""-"" + ""984F8"";\n    disp( SOL )\n    clear BAB;\n else\n    BAB = 1;\n    SOL = ""71E17"";\n end\n ')
end

// Thank you for the challenge!
CONCATENATED = exists( "BAB" );
if CONCATENATED == 1 then
   SOL = SOL + "-" + "984F8";
   disp( SOL )
   clear BAB;
else
   BAB = 1;
   SOL = "71E17";
end