c++ - Pass data as code -

i'm trying interpret formula given input:

y= y argv[1][s] 5; 

where argv[1][s] can + - * example.
y= y+5;
y= y*5;

i use check specific values, it's more interesting find out why doesn't work.

error c2146: syntax error : missing ';' before identifier 'argv'

i think happens + passed '+' no operation results. there way unquote this?

no, because that's not how c++ works. code must make sense @ compile-time, compiler can convert fixed set of assembler instructions. run-time text not "substituted" in; there no equivalent of "eval" in interpreted languages.

if want this, you'll need like:

switch (argv[1][s]) { case '+':     y = y + 5;     break; case '-':     y = y - 5;     break; case '*':     y = y * 5;     break; default:     std::cerr << "unrecognised operator: \"" << argv[1][s] << "\"" << std::endl;     break; }