使用算术操作符

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1 #!/bin/bash
2 # Counting to 11 in 10 different ways.
3
4 n=1; echo -n "$n "
5
6 let "n = $n + 1" # let "n = n + 1" 这么写也行
7 echo -n "$n "
8
9
10 : $((n = $n + 1))
11 # ":" 是必须的,这是因为,如果没有":"的话,Bash 将
12 #+ 尝试把"$((n = $n + 1))"解释成一个命令
13 echo -n "$n "
14
15 (( n = n + 1 ))

16 # 对于上边的方法的一个更简单的选则.
17 # Thanks, David Lombard, for pointing this out.
18 echo -n "$n "
19
20 n=$(($n + 1))
21 echo -n "$n "
22
23 : $[ n = $n + 1 ]
24 # ":" 是必须的,这是因为,如果没有":"的话,Bash 将
25 #+ 尝试把"$[ n = $n + 1 ]" 解释成一个命令
26 # 即使"n"被初始化成为一个字符串,这句也能工作.
27 echo -n "$n "
28
29 n=$[ $n + 1 ]
30 # 即使"n"被初始化成为一个字符串,这句也能工作.
31 #* Avoid this type of construct, since it is obsolete and nonportable.
31 #* 尽量避免这种类型的结果,因为这已经被废弃了,并且不具可移植性.
32 # Thanks, Stephane Chazelas.
33 echo -n "$n "
34
35 # 现在来个C 风格的增量操作.
36 # Thanks, Frank Wang, for pointing this out.
37
38 let "n++" # let "++n" also works.
39 echo -n "$n "
40
41 (( n++ )) # (( ++n ) also works.

42 echo -n "$n "
43
44 : $(( n++ )) # : $(( ++n )) also works.
45 echo -n "$n "
46
47 : $[ n++ ] # : $[ ++n ]] also works
48 echo -n "$n "
49
50 echo
51
52 exit 0
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