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Reference - What does this symbol mean in PHP?
I have this assignment:
$buffer=@data_value[$i];
what does the @ mean?
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15It's to prevent you from figuring out what's wrong with your code. – Ignacio Vazquez-Abrams Dec 08 '10 at 23:01
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1Actually that statement does not even compile – Matteo Riva Dec 08 '10 at 23:02
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@Ignacio I second that. Our phpcs code standards are configured to pick up any developers trying to do this and yell at them. – El Yobo Dec 08 '10 at 23:03
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@nico: `PHP Parse error: syntax error, unexpected '[' in...` – Matteo Riva Dec 08 '10 at 23:05
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Top tip - Only use the error suppressor when you absolutely have to - a reasonable example being a file operation you expect to fail, want to fail silently and intend to check yourself post-operation. – John Parker Dec 08 '10 at 23:05
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@nico - no, it doesn't, there's no $ on data_value. – El Yobo Dec 08 '10 at 23:06
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should be: `$buffer=@$data_value[$i];` – Stoic Dec 08 '10 at 23:19
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1@kemp: None of my PHP code compiles! – webbiedave Dec 08 '10 at 23:23
5 Answers
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The @ will suppress errors about variable not being initialized (will evaluate as null instead).
Also, your code is probably missing a $ after @:
$buffer=@$data_value[$i];
rustyx
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It is called the "error-control operator". Since it's an assignment, I believe you should do the rest yourself.
Jon
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the @ in front of a statement means that no warnings/errors should be reported from the result of that statement. To put simply, Error Reporting is suppressed for this statement.
This is particularly useful, when e.g. @fclose(fopen("file.txt",w")) which can throw several warnings/errors depending on the situation, but with an @ in front of it, all these warnings or errors will be suppressed.
Stoic
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Agreed. What I meant here was, this is useful when you are in a production environment and do not want to show error reports to end-user. – Stoic Dec 08 '10 at 23:18
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As above, it suppresses the error if the array key doesn't exist. A version which will do the same without resorting to dodgy error suppression is
$buffer = array_key_exists($i, $data_value) ? $data_value[$i] : null;
El Yobo
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That's just error suppression via syntactic salt, not a significantly different outcome. – mario Dec 08 '10 at 23:50
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No, it's correctly checking that a value exists before attempting to make use of it. – El Yobo Dec 09 '10 at 00:34