Problem with waiting for child processes within loop

[mhagger]

Configuration Information [Automatically generated, do not change]:
Machine: i386
OS: Linux
Compiler: gcc
Compilation CFLAGS:   -DHAVE_VFPRINTF -DHAVE_UNISTD_H -DHAVE_STDLIB_H 
-DHAVE_LIMITS_H        -DHAVE_GETGROUPS -DHAVE_RESOURCE -DHAVE_SYS_PARAM 
-DVOID_SIGHANDLER -DHAVE_GETDTABLESIZE -DHAVE_BCOPY -DHAVE_GETPW_DECLS 
-DHAVE_GETHOSTNAME -DHAVE_WAIT_H         -DHAVE_DUP2 -DHAVE_STRERROR  
-DHAVE_DIRENT_H -DHAVE_STRING_H        -DHAVE_VARARGS_H -DHAVE_STRCHR  
-DHAVE_DEV_FD        -Di386 -DLinux -DSHELL -DHAVE_ALLOCA -DHAVE_ALLOCA_H       
-O2 -march=i386 -mcpu=i686
uname output: Linux freak 2.2.14-12-custom #3 Mon May 15 00:12:59 EDT 2000 i686 
unknown

Bash Version: 1.14
Patch Level: 7

(I believe this is the copy of bash installed with RedHat Linux 6.2.)

Description:

I start multiple background processes, then try to wait for them in a
bash for loop.  One of the wait statements fails with the message
"wait: pid 2551 is not a child of this shell" when it should actually
succeed.

Repeat-By:

The following script produces the error on my machine:

    #! /bin/bash
    
    sleep 2 &
    pid1=$!
    echo "pid of task 1 is $pid1"
    
    sleep 1 &
    pid2=$!
    echo "pid of task 2 is $pid2"
    
    for pid in $pid1 $pid2
    do
        echo "waiting for $pid to finish"
        wait $pid
    done

I get the following output:

    $ ./wait-test
    pid of task 1 is 2550
    pid of task 2 is 2551
    waiting for 2550 to finish
    waiting for 2551 to finish
    ./wait-test: wait: pid 2551 is not a child of this shell

If I unroll the for loop (i.e., "wait $pid1; wait $pid2") or wait for
the two children in the same wait statement (i.e., "wait $pid1 $pid2")
then the error goes away.  If I change the second sleep statement to
"sleep 3" (i.e., so that the first child process finishes first), the
problem goes away.

Yours,
Michael

--
Michael Haggerty
mhagger@alum.mit.edu