为什么shell命令没有按我期望的顺序执行?

我写了3个名为s1.sh s2.sh s3.sh的shell脚本.它们具有相同的内容:

I have written 3 shell scripts named s1.sh s2.sh s3.sh. They have the same content:

#!/bin/ksh
echo $0 $$

和s.sh按顺序调用它们:

and s.sh invoke them in order:

#!/bin/sh
echo $0 $$
exec ./s1.sh &
exec ./s2.sh &
exec ./s3.sh &

但结果是混乱:

victor@ThinkPad-Edge:~$ ./s.sh
./s.sh 3524
victor@ThinkPad-Edge:~$ ./s1.sh 3525
./s3.sh 3527
./s2.sh 3526

为什么s1依次不是s2然后是s3?

why not s1 s2 then s3 in sequence?

如果我删除&在s.sh中:

If I remove & in s.sh:

#!/bin/sh
echo $0 $$
exec ./s1.sh 
exec ./s2.sh 
exec ./s3.sh 

输出:

$ ./s.sh
./s.sh 4022
./s1.sh 4022

为什么缺少s2和s3?

Missing s2 and s3, why?