In PHP, I haved tried this code print 08+"51";
but I don't know why it gives 51, while print 07+"51"; give 58 ?
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John Conde
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Van Minh Nhon TRUONG
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1See http://php.net/manual/en/language.types.integer.php: `$a = 0123; // octal number (equivalent to 83 decimal)` – Stephan Vierkant Jul 09 '18 at 14:31
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The upvote button states "this question shows research effort". No research effort is obvious yet it stands with 4 upvotes. – castis Jul 09 '18 at 14:34
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08 doesn't exist in octal – CSSBurner Jul 09 '18 at 14:54
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How did you not get Fatal Error? – CSSBurner Jul 09 '18 at 14:55
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@DavidPartyka: Because 08 is converted to 0, and "51" is converted to 51, so it gives 0+51 = 51 – Van Minh Nhon TRUONG Jul 09 '18 at 14:59
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Because when an integer starts with a 0 you are using octal. 08 is not a valid octal number so it translates to zero. 0 + 51 (because "51" is converted to a integer thanks to type juggling) equals 51.
John Conde
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1It should be "converted to an integer" not "converted to a string" -- it is already a string – johannes Jul 09 '18 at 14:32
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@johannes You are correct. I changed how I wanted to explain it halfway through and in the end got a bit of both. Fixed. – John Conde Jul 09 '18 at 14:52