shell - how to pass string to bash command as a parameter -

i have string variable contains loop.

loopvariable="for in 1 2 3 4 5 echo $i done" 

i want pass variable bash command inside shell script. getting error

bash $loopvariable 

i've tried

bin/bash $loopvariable 

but doesn't work. bash treats string giving me error. theoretically execute it. not sure doing wrong

bash: in 1 2 3 4 5 echo $i done: no such file or directory 

i have tried use approach using while loop. getting same error

i=0  loopvalue="while [ $i -lt 5 ]; make -j15 clean && make -j15 done" bash -c @loopvalue 

when use bash -c "@loopvalue" following error

bash: -c: line 0: syntax error near unexpected token `done' 

and when use use bash -c @loopvalue

[: -c: line 1: syntax error: unexpected end of file 

you can add -c option read command argument. following should work:

$ loopvariable='for in 1 2 3 4 5; echo $i; done' $ bash -c "$loopvariable" 1 2 3 4 5 

from man bash:

  -c         if -c option present, commands read               first non-option argument command_string.  if there argu‐              ments after command_string,    assigned                positional parameters, starting $0. 

another way use standard input:

bash <<< "$loopvariable" 

---

regarding updated command in question, if correct quoting issues, , fact variable not exported, still left infinite loop since $i never changes:

loopvalue='while [ "$i" -lt 5 ]; make -j15 clean && make -j15; done' i=0 bash -c "$loopvalue" 

but better use function in @kenavoz' answer.