PHP 5.3 and interface \ArrayAccess

Question

I'm now working on a project and I have one class that implements the ArrayAccess interface.

Howewer, I'm getting an error that says that my implementation:

must be compatible with that of ArrayAccess::offsetSet().

My implementation looks like this:

public function offsetSet($offset, $value) {
  if (!is_string($offset)) {
    throw new \LogicException("...");
  }
  $this->params[$offset] = $value;
}

So, to me it looks like my implementation is correct. Any idea what is wrong? Thanks very much!

The class look like this:

class HttpRequest implements \ArrayAccess {
  // tons of private variables, methods for working
  // with current http request etc. Really nothing that
  // could interfere with that interface.

  // ArrayAccess implementation

  public function offsetExists($offset) {
    return isset ($this->params[$offset]);
  }

  public function offsetGet($offset) {
    return isset ($this->params[$offset]) ? $this->params[$offset] : NULL;
  }

  public function offsetSet($offset, $value) {
     if (!is_string($offset)) {
      throw new \LogicException("You can only assing to params using specified key.");
     }
     $this->params[$offset] = $value;
  }

  public function offsetUnset($offset) {
    unset ($this->params[$offset]);
  }
}

The class look like this:

class HttpRequest implements \ArrayAccess {
  // tons of private variables, methods for working
  // with current http request etc. Really nothing that
  // could interfere with that interface.

  // ArrayAccess implementation

  public function offsetExists($offset) {
    return isset ($this->params[$offset]);
  }

  public function offsetGet($offset) {
    return isset ($this->params[$offset]) ? $this->params[$offset] : NULL;
  }

  public function offsetSet($offset, $value) {
     if (!is_string($offset)) {
      throw new \LogicException("You can only assing to params using specified key.");
     }
     $this->params[$offset] = $value;
  }

  public function offsetUnset($offset) {
    unset ($this->params[$offset]);
  }
}

The line with class declaration (class X implements \ArrayAccess). — Jakub Lédl
– Jakub Lédl, Commented Feb 13, 2010 at 13:33
It could help if you provide the rest of the class (or a shortened version of it). — Lior Cohen
– Lior Cohen, Commented Feb 13, 2010 at 13:33
Is your class declaration inside a namespace? e.g. do you have a namespace My\Namespace; declaration at the top of the file? if not, you shouldn't need to qualify ArrayAccess. You could also try "use"ing ArrayAccess instead of fully qualifying it. This seems like a weird error, so I'm just throwing things out there... — tmont
– tmont, Commented Feb 14, 2010 at 3:00

Kris · Accepted Answer · 2010-06-14 12:18:32Z

1

Looks to me like your namespace or use directives in the top of the file make it look for the wrong ArrayAccess interface to be compatible with. Can't tell for sure without those directives though.

In general:

Your own namespaces should not begin or end with a backslash:

Use: namespace Web\Http;

Don't use something like: namespace \Web\Http; or namespace \Web\Http\;

For every class and interface you reference in your file, add a use directive:

namespace MyProject;

use MyLibrary\BaseClass; // note no backslash before the namespace name
use \ArrayAccess;
use \Iterator;
use \Countable;

class MyClass extends BaseClass implements ArrayAccess, Iterator, Countable
{
    /* Your implementation goes here as normal */
}

answered Jun 14, 2010 at 12:18

Kris

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Comments

Lior Cohen · Accepted Answer · 2010-02-13 13:55:12Z

0

The only thing that catches my eye here:

 public function offsetGet($offset) {
    return isset ($this->params[$offset]) ? $this->params[$offset] : NULL;
  }

Perhaps replacing it with:

 public function offsetGet($offset) {
    return (isset ($this->params[$offset]) ? $this->params[$offset] : NULL);
  }

would get the trick done.

It could also be a syntax error that drags on from the part of the code you haven't pasted.

edited Feb 13, 2010 at 13:55

answered Feb 13, 2010 at 13:49

Lior Cohen

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6 Comments

Jakub Lédl Over a year ago

Thanks, but although this is surely correct and it improves the code readability, it doesn't help me with this problem.

Jakub Lédl Over a year ago

Good idea with that syntax error, but after removing the implementation the class works perfectly again.

Lior Cohen Over a year ago

Your code works fine here. Pasted it into a file and ran it from the command line using: php test.php. It has to be something in the part of the code you haven't pasted.

Lior Cohen Over a year ago

Hey, what version of PHP are you using? PHP 5.3.X?

Jakub Lédl Over a year ago

I've just did the same, with the implementation commented out, and it worked. The code above is fine. It should be fine anyway. I don't say that Netbeans 6.8 IDE is completely reliable, but it doesn't report any syntax error.

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