CODEWITHSUNDEEP
perl
NeilWang
NeilWang

Reputation: 367

What happened in autoincrement function in Perl?

I've got a question in Perl studying.

$a = "Z9"; print ++$a;

It prints out AA0. This can easily understood that 9 increased by 1 gets 10 and Z increased 1 gets AA. Just like 99+1. But what happened if I flipped over Z and 9:

$a = "9Z"; print ++$a;

This will get 10. Why the letter Z is gone? Where did it gone?

Regards, Neil

Upvotes: 1

Views: 135

Answers (1)

Miller
Miller

Reputation: 35208

"9Z" is interpreted like a number, and therefore the non-numeric ending gets truncated before the autoincrement. The same would be true of "9 the rest of this string is ignored".

This is similar to the way that print "2foo" + "2bar"; will output 4.

The string magic that you observe with "Z9" is documented in perlop #Auto-increment and Auto-decrement

The auto-increment operator has a little extra builtin magic to it. If you increment a variable that is numeric, or that has ever been used in a numeric context, you get a normal increment. If, however, the variable has been used in only string contexts since it was set, and has a value that is not the empty string and matches the pattern /^[a-zA-Z]*[0-9]*\z/ , the increment is done as a string, preserving each character within its range, with carry:

  1. print ++($foo = "99"); # prints "100"
  2. print ++($foo = "a0"); # prints "a1"
  3. print ++($foo = "Az"); # prints "Ba"
  4. print ++($foo = "zz"); # prints "aaa"

Upvotes: 5

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