Question: How do you insert a printf comment inside of an echo? I have been trying to figure it out for the past hour and everything I try results in a different error.

Below is the last version of the code that I tried.

```
<?php
$i = 0;
do {
    echo "<li><a class=\"th radius\" href=\"img/coolstuff/" . printf("%03d", $i); . ".jpg\"><img src=\"img/coolstuff/" . printf("%03d", $i); . ".jpg\"></a></li>";
    $i++;
} while ($i < 282);
?>

```

Answer: 
Remove the `;` before the `.`, otherwise you're terminating the statement.

Additionally, `printf` directly outputs the string, which will make it appear at the *start* of the outputted string. You want `sprintf` instead.
Do like this:

```
echo "<li><a class=\"th radius\" href=\"img/coolstuff/" , printf("%03d", $i);

```

Notice the comma `,` instead of dot `.`.

When using a comma with a `echo` statement, like above, each part is evaluated first. Atleast according to [this accepted anwer.](https://stackoverflow.com/questions/4393411/what-does-comma-in-echo-statement-signify)