argument.printf

Code example

<?php declare(strict_types = 1);

$name = 'world';
printf('%s is %d years old', $name);

Why is it reported?

The number of format placeholders in the printf format string does not match the number of values passed. printf requires exactly as many value arguments as there are placeholders in the format string. A mismatch will result in missing values or ignored extra arguments at runtime.

In the example above, the format string '%s is %d years old' contains 2 placeholders (%s and %d), but only 1 value ($name) is provided.

How to fix it

Pass the correct number of arguments to match the placeholders:

 <?php declare(strict_types = 1);

 $name = 'world';
-printf('%s is %d years old', $name);
+printf('%s is %d years old', $name, 25);

Or adjust the format string if fewer values are intended:

 <?php declare(strict_types = 1);

 $name = 'world';
-printf('%s is %d years old', $name);
+printf('%s says hello', $name);