Using exprtk, what is the correct way to create a vector based variable from within an unknown_symbol_resolver?

Question

The exprtk documentation provides a great example on how to create a scalar variable from an unknown_symbol_resolver. How to do I create a vector? I would expect to find a create_vector member function on the symbol_table class, but there is none.

Consider a simple program that takes an exprtk formula from the command line. In my example, I'd like to have foobar created as a variable instead of a scalar.

./a.out "return [ foobar ]; "

The program.

#include <string> #include <stdio.h> #include <vector> #include "exprtk.hpp" using symbol_table_t = exprtk::symbol_table<double>; using parser_t = exprtk::parser<double>; using expression_t = exprtk::expression<double>; template<typename T> struct myusr : public parser_t::unknown_symbol_resolver { using usr = typename parser_t::unknown_symbol_resolver; myusr() : usr(usr::e_usrmode_extended) { } virtual bool process(const std::string &symbol, symbol_table_t &symbol_table, std::string &) { printf("Handling symbol: %s\n", symbol.c_str()); symbol_table.create_variable(symbol, 4.14); return true; } }; int main(int argc, char **argv) { symbol_table_t symbol_table; symbol_table.add_constants(); expression_t expression; expression.register_symbol_table(symbol_table); parser_t parser; myusr<double> lookup; parser.enable_unknown_symbol_resolver(&lookup); if (!parser.compile(argv[1], expression)) { printf("Compilation error: %s\n", argv[1]); return -1; } return 0; }

Rafael · Accepted Answer · 2025-04-17 10:34:12Z

The answer appears to be that rather than expect a functional create_vector member function, one can just use add_vector.

However, its critical to provide a vector with at least one element and the vector must have storage beyond the lifetime of the invocation of the unknown_symbol_resolver::process member function.

By way of an example. One could write it like this.

 virtual bool process(const std::string &symbol, symbol_table_t &symbol_table, std::string &) { static std::vector<T> x; x.push_back(3.14); symbol_table.add_vector(symbol, x); return true; }

Of course, this example is in some ways terrible, because the static scoped storage for x would be shared for every unknown variable.

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