How do I evaluate strings to numbers in Bash?

It's like Linux Everything in is a file ,bash Everything in is a string .

Technically speaking ,Bash There is no data type in , Essentially ,Bash Variables are just strings .

When you try in bash When performing arithmetic operations in , This can cause problems , The number you're trying to add will give you strange results .

ywmt@wljslmz:~$ sum=3+6
ywmt@wljslmz:~$ echo $sum
3+6

To be in bash Convert string to integer , Please pack them in $((X)) in , like this ：

$((string))

If the string has a recognizable number , It will be treated as a number , The example I showed above can be corrected to ：

ywmt@wljslmz:~$ sum=$((3+6))
ywmt@wljslmz:~$ echo $sum
9

You are in $((…)) Anything done in is considered an arithmetic operation .

Let's look at string to number conversion in more detail , So that you can go to bash Use arithmetic calculations in .

Convert a string variable to a number

Let's declare something that is basically a string “ Numbers ”.

ywmt@wljslmz:~$ a=11
ywmt@wljslmz:~$ b=3

You can check the assigned number ：

ywmt@wljslmz:~$ echo $a
11
vimanyu@wljslmz:~$ echo $b
3

Now let's try adding the two numbers and storing the value in the third variable c：

ywmt@wljslmz:~$ c=$a+$b
ywmt@wljslmz:~$ echo $c
11+3

As you can see in the output above ,c It is also regarded as a string .

To make the addition successful , You need to expand by arithmetic “ transformation ” it ：

ywmt@wljslmz:~$ c=$(($a+$b))
ywmt@wljslmz:~$ echo $c
14

Please note that ,c It's actually still a string , Until you implicitly use the same arithmetic extension again for another operation I'll discuss later .

ywmt@wljslmz:~$ c=$((5))

This is equivalent to ：

ywmt@wljslmz:~$ c=5

d Let's use the third variable for another operation ：

ywmt@wljslmz:~$ d=10

All these implicit declarations are still strings . This is different from the way I defined before a identical b.

Now proceed to the second operation ：

ywmt@wljslmz:~$ e=$(($a+$b*$c-$d))
ywmt@wljslmz:~$ echo $e
16

In the above expression , First calculate b and c The product of the , Finally, calculate addition and subtraction .

As I mentioned before , You are in $((…)) Anything done between is considered arithmetic .

All these conversions apply to integers . It does not apply to floating point numbers , A number with a decimal point .

Try mixing actual strings and numbers

Mixing strings and integers is still safer than mixing coke and mint . however , It does produce a strange result .

ywmt@wljslmz:~$ sum=$((3+hello))
ywmt@wljslmz:~$ echo $sum
3

It's only when there are numbers （ As string ） Only when effective , If it's a pure string , It won't magically convert into some numbers , contrary , It will be ignored , As you can see in the example above .

Alternative methods ： Use expr

You can also use this expr Tools to evaluate , But please note that it is not “ This machine ” Bash Program , Because you need coreutils install （ By default, in Ubuntu On ） As a separate package .

ywmt@wljslmz:~$ expr $a + $b \* $c - $d
16