How To Increment A Variable In Bash

Scripting is deeply integrated into bash. For example, I’d written earlier on how to efficiently concatenate strings. In the same way, you can use it as a programming language to perform various tasks, and this means that there are different ways to increment a variable in bash. In this article, I’ll show you the different ways to do this along with a comparison of what separates the various techniques from each other.

So let’s get started!

Different Methods to Increment a Variable in Bash

Here are all the different ways you can increment a bash variable. Some of these are shorter than others but come with the disadvantage of being harder to understand. So you should choose the one that best suits your needs. In addition, there are efficiency considerations that might play a role.

Using the “Let” Command

The “let” command in bash evaluates arithmetic expressions. Those of you who are familiar with the BASIC programming language might get the impression that it’s the same. Still, the purposes of the statement in the two languages are very different. In bash scripting, we use let whenever we want to perform a mathematical operation.

In the context of incrementing a variable in bash, we can use “let” like this:

#!/bin/bash
x=5
echo "Original value: $x"
let x=x+1
echo "Incremented value: $x"

This script will first print “5”, and then print “6” as shown here:

Note that in the screenshot, I’m separating the statements via a semi-colon (;) since I’m entering the commands directly into the bash prompt instead of through a script. You can see the statement:

let x=x+1

This is the line that increments the value of the variable.

Arithmetic Expansion

Arithmetic Expansion is a tool in the bash scripting language, that allows you to evaluate a result and then substitute that result in place of the expression. It has the following syntax:

$((expression))

For incrementing a variable, there are two ways you can use arithmetic expansion, depending upon whether you want to use the post/pre-fix increments like “x++”. If you want to use increments, you can do it like this:

#!/bin/bash
x=5
echo "Original value: $x"
((x++))
echo "Incremented value: $x"

In the above code, the following line is the arithmetic expansion:

((x++))

You can also use the following:

#!/bin/bash
x=5
echo "Original value: $x"
x=$(($x + 1))
echo "Incremented value: $x"

Here, instead of using an increment operator, we use a plain arithmetic expression:

x=$(($x + 1))

In other programming languages, you don’t need to have this kind of weird syntax just to perform basic operations, but that’s how it is in bash.

Using “expr”

“Expr” is a way to spawn a new process for incrementing a variable, and is thus the least efficient of these above methods. Here’s what it looks like in action:

#!/bin/bash
x=5
echo "Original value: $x"
x=$(expr$x + 1)
echo "Incremented value: $x"

There was a time when bash didn’t have the newer operators like $(( )), and for these older systems, using “expr” allows for code to be run on older machines. But as they vanish, these are getting phased out so its use is becoming rarer and rarer, particularly for something as simple as just increasing the value of a variable by one.

Why Is Bash So Convoluted When It Comes To Incrementing Variables?

If you’re used to other programming languages like Java or Perl, you might be puzzled as to why the bash scripting language has such weird rules for simple arithmetic operations. In other languages, you just need to type the expression, assign it to a variable, and you’re done. For example, in C, you just need to do this:

x++;

No need for special operators or “let” statements. So why was it designed like this?

Bash Was Designed for Text

Bash and scripting were designed to automate server management tasks, so the primary focus is on text manipulation and strings. As a result, the string manipulation capabilities are front and center, and arithmetic operations are secondary. I don’t know if they would design it this way if they were to do it all over again, but it explains why it’s so weird to increment a variable in Bash, when other languages are simpler.

Lack of Data Types

Related to the first point, Bash tends to treat variables as strings. So simply incrementing a numerical variable doesn’t immediately make sense unless the context is specifically an arithmetic operation – which is why you need a declarative statement like “let”, or an arithmetic expansion operator like the double brackets (( )).

Compatibility with Older Code

Often, design decisions in the early stage of a programming language can last for years. If you want older code to continue to run, then you don’t have a choice but to let the syntax stay the way it is. And by now, it’s entirely too late. There’s so much legacy code, that it’s easier to keep things as they are than to rewrite a scripting language that everyone’s already familiar with.

Safety

Since bash scripts are widely used for server automation, there’s an increased emphasis on safety. Because of this, Bash script is designed to ensure that there can be no mistaking what a user intends. For example, some programming languages like JavaScript allow you to use the plus sign (+) to concatenate strings! If you want to treat a string as a number, you have to explicitly inform the system that you’re treating it as a number.

In a server environment, this can raise all kinds of security issues – both accidental, as well as malicious in the form of code injections. So the default approach is to prioritize safety and require scripts to use special notations when they want to increment a variable in Bash.

Prefix and Postfix Increments for Incrementing a Variable

There are two quick ways of incrementing a variable in Bash. Prefix and postfix increments. They look like this:

1. Prefix increment: ++x
2. Postfix increment: x++

Of these, the postfix increment is more widely known, but in reality, you’ll often want to use the prefix one!

The difference is this. In the prefix increment, the system increases the variable by one and returns the value. In the postfix increment, the system first returns the value and then increases it by one.

Here’s an example of some code that illustrates this:

x=5
echo "Before prefix increment: x=$x"
echo "Using prefix increment: ++x="$((++x))
echo "After prefix increment: x=$x"
x=5
echo "Before postfix increment: x=$x"
echo "Using postfix increment: x++="$((x++))
echo "After postfix increment: x=$x"

And here’s what the output looks like:

In this code, I first display the value of a variable, then use an “echo” statement to display the prefix and postfix operation. You can see that both the before and after values are the same. However, while using the expression in a statement, the prefix increment first increases the value and then returns it, while the postfix expression does the reverse.

The difference between these two is subtle, and it’s particularly important to be aware of it when using the incremented value in for-loop blocks.

Conclusion

There are several ways to increment the value of a variable in Bash. Hopefully, the above examples give you a comprehensive understanding of how to do this and the nuances of prefix and postfix increments.