How can I check the length of a string in a Bash loop at a specific iteration?

Here’s a clean approach:

```bash
#! /bin/bash
var="nef892na9s1p9asn2aJs71nIsm"
for counter in {1..40}; do
var=$(printf '%s' "$var" | base64)
if [ $counter -eq 35 ]; then
printf '%d' ${#var}
fi
done
```

This avoids adding a newline every time, and you should only see the count at iteration 35.

**Key Takeaway:** Use `printf` instead of `echo` to prevent newline issues. Here’s a simple script you can run:

```bash
#! /bin/bash
var="nef892na9s1p9asn2aJs71nIsm"
for counter in {1..40}; do
var=$(printf '%s' "$var" | base64)
if [ $counter -eq 35 ]; then
echo -n ${#var}
break
fi
done
```

This will print just the character count at iteration 35 ready for you to use!

This code should work for you:

```bash
#! /usr/bin/env bash
var="nef892na9s1p9asn2aJs71nIsm"
for counter in {1..40}; do
var=$(echo $var | base64 --wrap=0)
[[ $counter == 35 ]] && echo -n ${#var}
done
```

The `--wrap=0` prevents any wrapping issues, so you should just get the number without newline characters.

You might want to use something like this:

```bash
for ((counter = 1; counter <= 40; counter++)); do
currentString=$(printf '%s' "$currentString" | base64 | tr -d 'n')
if [[ $counter -eq 35 ]]; then
printf '%dn' "${#currentString}"
break
fi
done
```

Make sure to use `printf` instead of `echo`, as it helps avoid issues with newlines that can mess up your count.