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author | Herbert Xu <herbert@gondor.apana.org.au> | 2009-02-22 18:10:01 +0800 |
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committer | Herbert Xu <herbert@gondor.apana.org.au> | 2009-02-22 18:10:01 +0800 |
commit | 3800d4934391b144fd261a7957aea72ced7d47ea (patch) | |
tree | 40c003ab3063ceab7f3615a623a09d3c610332a0 /src/exec.h | |
parent | Release 0.5.5.1. (diff) | |
download | dash-3800d4934391b144fd261a7957aea72ced7d47ea.tar.gz dash-3800d4934391b144fd261a7957aea72ced7d47ea.zip |
[JOBS] Fix dowait signal race
This test program by Alexey Gladkov can cause dash to enter an infinite loop in waitcmd. #!/bin/dash trap "echo TRAP" USR1 stub() { echo ">>> STUB $1" >&2 sleep $1 echo "<<< STUB $1" >&2 kill -USR1 $$ } stub 3 & stub 2 & until { echo "###"; wait; } do echo "*** $?" done The problem is that if we get a signal after the wait3 system call has returned but before we get to INTON in dowait, then we can jump back up to the top and lose the exit status. So if we then wait for the job that has just exited, then it'll stay there forever. I made the original change that caused this bug to fix pretty much the same bug but in the opposite direction. That is, if we get a signal after we enter wait3 but before we hit the kernel then it too can cause the wait to go on forever (assuming the child doesn't exit). In fact this is pretty much exactly the scenario that you'll find in glibc's documentation on pause(). The solution is given there too, in the form of sigsuspend, which is the only way to do the check and wait atomically. So this patch fixes Alexey's race without reintroducing the old bug by converting the blocking wait3 to a sigsuspend. In order to do this we need to set a signal handler for SIGCHLD, so the code has been modified to always do that. Signed-off-by: Herbert Xu <herbert@gondor.apana.org.au>
Diffstat (limited to 'src/exec.h')
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