2

I've made something that I think ought to be a shell script called bashispainful.sh:

Code:

#!/bin/bash Var1= $((10*5)) Var2= $((10/5)) Var3= $((10-5)) Var4= $((10+5)) Var5= ($(Var4)*$(Var3)) echo $Var1 echo $Var2 echo $Var3 echo $Var4 echo $Var5

I would like to receive the output:

50 2 5 15 75

My current output:

./Bashispainful.sh: line 2: 50: command not found ./Bashispainful.sh: line 3: 2: command not found ./Bashispainful.sh: line 4: 5: command not found ./Bashispainful.sh: line 5: 15: command not found ./Bashispainful.sh: line 6: syntax error near unexpected token `(' ./Bashispainful.sh: line 6: `Var5= ($(Var4)*$(Var3))'

My aim here is to get any kind of grip on being able to do basic maths in bash. I have tried every website and every combination of brackets, dollar signs and lack thereof that I can think of and frankly I am baffled and desparate. How can I perform simple maths in bash?

(Please make this easy to understand - please assume that I am a non-programmer and stupid)

Improve this question
1
  • By the way, if you are doing arithmetic, you can use the ((...)) arithmetic form, in which variables don't need to be prefixed with a $ (unless they are positional parameters like $3) and spaces don't matter. For example: (( a = 42 , b = a - 27)) Commented Nov 23, 2014 at 3:06

4 Answers 4

Reset to default
4

Bash is not painful:

#!/bin/bash Var1=$((10*5)) Var2=$((10/5)) Var3=$((10-5)) Var4=$((10+5)) Var5=$((Var4*Var3)) echo $Var1 echo $Var2 echo $Var3 echo $Var4 echo $Var5

The output is:

50 2 5 15 75

Notes

Consider:

Var1= $((10*5))

After bash does arithmetic substitution, that command looks like:

Var1= 50

That command says (a) temporarily set Var1 to empty, and (b) run the command named 50. Since there is no command named 50, you receive the error:

Bashispainful.sh: line 2: 50: command not found

The solution is remove the space so that the Var1 is assigned to a non empty value:

Var1=$((10*5))

Consider:

Var5= ($(Var4)*$(Var3))

This sets Var5 temporarily to empty for a command whose name starts with (. That is not valid. Hence the error:

./Bashispainful.sh: line 6: syntax error near unexpected

The correct bash syntax for arithmetic is $((...)). Thus, that line could be written:

Var5=$(($Var4*$Var3))

However, because bash is not painful, it understands that strings inside arithmetic expressions are shell variables. Thus, the dollar signs are unnecessary and the above can be simplified to:

Var5=$((Var4*Var3))

Alternatively, since the above is just doing assignment to variable, it can be further simplified to:

((Var5=Var4*Var3))

Note that $((...)) returns the value of the arithmetic computation but ((...)) does not return any value. It does set an exit code which makes it useful for tests such as used in if statements.

Improve this answer
Sign up to request clarification or add additional context in comments.

6 Comments

So what you are saying is that spaces matter in bash? (Never used a language where they even slightly mattered before) Bash IS going to be painful. Thank you nonetheless. Definitely an upvote!
Yes, spaces matter in bash, as they do in any other shell. If you think this is painful, consider space a character like every other: why should an x matter, or a quotation mark, but not a whitespace character?
I see - I'm really going to have a hard time with this.
@Reluctant_Linux_User Computer languages are each designed for a different purpose. Accept it for what it is and bash is quite powerful: Var= command executes a command with Var set to empty while Var=command sets the variable to have the value command.
@Reluctant_Linux_User Not that I am aware of and certainly not one that meets the POSIX standard.
|
2

Remove the space before the $ and move the bracket from the line in the Var5 declaration.

#!/bin/bash Var1=$((10*5)) Var2=$((10/5)) Var3=$((10-5)) Var4=$((10+5)) Var5=$((Var4*Var3)) echo $Var1 echo $Var2 echo $Var3 echo $Var4 echo $Var5

The space screws with the declarations, and you want to read the value from Var4 and Var3, so you need two brackets.

Improve this answer

Comments

0

You can't put spaces before or after the =, or bash tries to execute the command $((10*5)) (i.e., 50) while setting Var1 to the empty string in a local environment. Also, using single parentheses without a dollar sign (…) invokes a subshell, with a dollar sign $(…) invokes a subshell and is substituted with the output of a command, and the dollar sign is optional inside an arithmetic expression.

So what you want to do is:

#!/bin/bash Var1=$((10*5)) Var2=$((10/5)) Var3=$((10-5)) Var4=$((10+5)) Var5=$((Var4*Var3)) echo $Var1 echo $Var2 echo $Var3 echo $Var4 echo $Var5
Improve this answer

2 Comments

You can omit the preceding $ from variables inside $((...)). OK you did in your edit..
Saw that, good thing is that they are essentially optional for regular variables. It will work either way. They are only required when including a brace protected variable (e.g. ${var:(-3):2}).
0

check bash command "expr"

For example, if you wnat to add two number: var=expr 2 + 6

Please refer to the man page for more expression.

Improve this answer

4 Comments

What is the "Man page"? I am a serious noob.
you can type "man expr" in bash. Or here is a webpage for expr" ss64.com/bash/expr.html
expr is unneccessarily verbose. Why not use $(())?
Hello and welcome to StackOverflow. Could you please format this question so it is cleaner and expresses your intent more clearly, along with helpful explanation as to what is going on?

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.