Question: I have 3 text files. I would like to read them and store them in different variables and later concatenate them using paste and print them in console.
I tried the following code but it threw an error saying 

> 
> File not found
> 
> 
> 

Here is my code

```
#!/bin/sh
value_1=`cat file_1.txt`
value_2=`cat file_2.txt`
value_3 = paste $value_1 $value_2
echo "$value_3"

```

Answer: 
`paste` expects its arguments to be the names of files, not the content of files. With bash, ksh, or zsh, there is a way around this. Replace:

```
paste $value_1 $value_2

```

with:

```
paste <(echo "$value_1") <(echo "$value_2")

```

`<(...)` is called *process substitution*. It makes the output from the command inside the parens look like a file.

### Improvement

If we don't know the first character in the output, then `printf` is more reliable than `echo`:

```
paste <(printf "%s" "$value_1") <(printf "%s" "$value_2")

```

### Example

Let's use these two test files:

```
$ cat file1
1
2
$ cat file2
a
b

```

Now, let's read those files into variables and apply paste to those variables:

```
$ value_1=$(cat file1); value_2=$(cat file2)
$ paste <(printf "%s" "$value_1") <(printf "%s" "$value_2")
1       a
2       b

```

Or, saving the output in a variable:

```
$ value_3=$(paste <(printf "%s" "$value_1") <(printf "%s" "$value_2"))
$ echo "$value_3"
1       a
2       b

```
The third line should read

```
value_3=`paste file_1.txt file_2.txt`

```

You need the backticks, no space after `value_3` and don't use the variables as arguments, use the file names.

The reason it is saying "file not found" is because the `value_3` with a space after it is being interpreted as a command to be run.