java - Writing 2^x in PL -

we designed pl has following syntaxes:

load variablename, value inc variablename //increases value 1 loop variablename //loops number of times depending                     on variablename's value @ time. if                     variablename's value somehow altered later                     on in code, loop cycle runs same                     amount variablename's default value. end //we put corresponding end our loop. 

for example here simple code computes x+x:

vm vm6 = new vm();             vm6.add("load x, 7");             vm6.add("load answer, 0");             vm6.add("loop x");             vm6.add("inc answer");             vm6.add("inc answer");             vm6.add("end"); 

now trying figure out how write 2^x stuck on longest time. me?

edit: figured out. used own solution of nested loops.

it seems pretty straightforward, doesn't it? compute x^y, loop y times, each time multiply answer x. answer starts off being 1 (anything 0 power 1).

load x, 2 load answer, 1 loop [power] [multiply procedure] end 

the [multiply procedure] referred above additive method of multiplication. 2 numbers multiplied (x * y) equal number of intersections between 2 loops of length x , y. purposes, multiplying x * y looks this:

load x, [x value] load y, [y value] load answer, 0 loop x loop y inc answer end end 

so whole 2^x procedure series of loops:

load x, [power] load base, 2 load answer, 1 loop x     load addition, 0 //reset multiplier     loop answer         loop base             inc addition         end     end     load answer, addition // load answer multiplied value end 

of course, if can't nested loops, you're pretty out of luck.