posix - Is my shell broken? -

#!/bin/sh count=0 foo=0  echo "foo $foo"  while [ "$foo" -eq 0 ] && [ "$count" -lt 10 ];     echo "inside while: foo $foo"     count=$((count+1))     foo=1 done  echo "after while" 

you'd expect script above output following, right?

foo 0 inside while: foo 0 after while 

and on many other machines, own. not mine ...

foo 0 inside while: foo 0 inside while: foo 1 inside while: foo 1 ... (infinite loop) 

am doing wrong, or missing obvious?

on (broken?) machine, if tweak while conditional use "obsolescent" -a operator, "fixes" problem (whatever is). why?

assuming no hidden characters in script text (an assumption suggest putting effort verifying!), indeed exhibiting behavior contrary posix sh standard.

n1 -eq n2 - true if integers n1 , n2 algebraically equal; otherwise, false.

notably, makes no specifications or guarantees happens if either argument not in fact integer; instance, if would be integer, has carriage return or other nonprinting character on end.

other items try, in order:

• run sh -x yourscript, , @ exact arguments called test when running loop. determine whether second test in && chain run @ all.
• change operator -eq =, running string comparison rather numeric comparison (and thereby ensuring string containing hidden characters fail compare string 0, rather relying on undefined behavior in case).
• replace [ "$foo" -eq 0 ] false, , ensure contents of loop not run (thus sanity-checking other core parts of shell's implementation).