On Wed, Mar 19, 2014 at 02:18:52PM +0400, Dmitry Arkhireev wrote: > And if run it sample with echo $I after ((I++)) everything works as expected > I=0; while [ $I -ne 1 ]; do ((I++)); echo $I; done; echo $? > 1 > 0
Here, $? is the exit status of the echo, instead of the ((...)) command. The ((...)) command returns a status of either 0 or 1, depending on the value of the expression it evaluates. With C-style post-increment (++) the expression evalutes to the original value of the variable, rather than the incremented value. imadev:~$ i=0; ((i++)); echo $? 1 imadev:~$ i=1; ((i++)); echo $? 0 imadev:~$ i=999; ((i++)); echo $? 0 When the original value is 0, the expression evaluates to "false" (using the semantics of C), and "false" causes an exit status of 1 (using the semantics of Bash). When the original value is anything other than 0, the expression evaluates to "true" (C-style), which causes an exit status of 0 (Bash-style).
