如果在 bash 中进行语句算术，我该怎么做？
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How do I do if statement arithmetic in bash?
提问by jmasterx
I want to do something like this:
我想做这样的事情：
if [ % 4 == 0 ]; then
...
But this does not work.
但这不起作用。
What do I need to do instead?
我需要做什么？
回答by guns
read n
if ! ((n % 4)); then
echo "$n divisible by 4."
fi
The (( ))operator evaluates expressions as C arithmetic, and has a boolean return.
该(( ))运算符将表达式计算为 C 算术，并返回布尔值。
Hence, (( 0 ))is false, and (( 1 ))is true. [1]
因此，(( 0 ))是假的，(( 1 ))是真的。[1]
The $(( ))operator also expands C arithmetic expressions, but instead of returning true/false, it returns the value instead. Because of this you can test the output if $(( ))in this fashion: [2]
该$(( ))运算符还扩展了 C 算术表达式，但它不返回真/假，而是返回值。因此，您可以$(( ))以这种方式测试输出：[2]
[[ $(( n % 4 )) == 0 ]]
But this is tantamount to: if (function() == false). Thus the simpler and more idiomatic test is:
但这无异于：if (function() == false)。因此，更简单和更惯用的测试是：
! (( n % 4 ))
[1]: Modern bash handles numbers up to your machine's intmax_tsize.
[1]：现代 bash 可以根据您的机器intmax_t大小处理数字。
[2]: Note that you can drop $inside of (( )), because it dereferences variables within.
[2]：请注意，您可以$放入 of 内(( ))，因为它会取消引用其中的变量。
回答by Jan Vorcak
a=4
if [ $(( $a % 4 )) -eq 0 ]; then
echo "I'm here"
fi
回答by chown
single brackets ([..]) don't work for some tests, try with double brackets ([[...]]) and enclose the mod in ((..))to evaluate the %operator properly:
单括号 ( [..]) 不适用于某些测试，请尝试使用双括号 ( [[...]]) 并将 mod 括起来((..))以%正确评估运算符：
if [[ $(( % 4 )) == 0 ]]; then
More details here:
http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_02.html
更多细节在这里：http:
//tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_02.html
回答by potong
This might work for you:
这可能对你有用：
((a%4==0)) && echo "$a is divisible by 4" || echo "$a is not divisible by 4"
or more succinctly:
或更简洁地说：
((a%4)) && echo "$a is not divisible by 4" || echo "$a is divisible by 4"
回答by bballdave025
If you want something a bit more portable - for example, something that works in shas well as in bash- use
如果你想要一些更便携-例如，一些作品中sh，以及在bash-使用
if [ $(echo " % 4" | bc) -eq 0 ]; then
...
An example
一个例子
#!/bin/bash
#@file: trymod4.bash
if [ $(echo " % 4" | bc) -eq 0 ]; then
echo " is evenly divisible by 4"
else
echo " is NOT evenly divisible by 4"
fi
$ chmod +x trymod4.bash
$ ./trymod4.bash 224
224 is evenly divisible by 4
$ ./trymod4.bash 223
223 is NOT evenly divisible by 4
I put this in, because you used the single [ ... ]conditional, which I usually associate with sh-compatible programs.
我把它放进去，因为你使用了单个[ ... ]条件，我通常将它与sh兼容程序相关联。
Check that this works in sh.
检查这在sh.
#!/bin/sh
#@file: trymod4.sh
if [ $(echo " % 4" | bc) -eq 0 ]; then
echo " is evenly divisible by 4"
else
echo " is NOT evenly divisible by 4"
fi
$ chmod +x trymod4.sh
$ ./trymod4.sh 144
144 is evenly divisible by 4
$ ./trymod4.sh 19
19 is NOT evenly divisible by 4
All right, it works with sh.
好吧，它适用于sh.
Note the "theoretical" (but not always implemented as such) differences between [ ... ]and [[ ... ]]from this site(archived).
