Chet Ramey a écrit : > On 3/4/10 2:36 AM, Ettelbrueck, Heiko wrote: >> Problem: The $? variable is always 0 after that statement. (If, >> on the other hand, I separate the declaration and the >> definition of the variable as shown in the example below, the >> $? variable is really set to the exit status of the external >> tool.) > > The exit status is the status of the command you run: local. In the > absence of a command, when there are only assignment statements, the > exit status can be the exit status of a command substitution:
I have been bitten by this a few times. Since then I never, ever write this: local foo=... But always: local foo; foo=... It's overkill, but simple. I tend to even forget why I am doing this, which is nice: now I can focus on the real problems to solve.
