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Postings from January 2016
Am 27.01.2016 um 17:32 schrieb Wolf-Dietrich Moeller (via RT): > # New Ticket Created by Wolf-Dietrich Moeller > # Please include the string: [perl #127391] > # in the subject line of all future correspondence about this issue. > # <URL: https://rt.perl.org/Ticket/Display.html?id=127391 > > > > Hello, > I am not sure if the following issue in the "perlop"-documentation needs > clarification. > > In the section on "Operator Precedence and Associativity" there is the line: > "right = += -= *= etc. goto last next redo dump" > The "etc." means to me that all other assignment operators are also > right-associative, i.e. in particular also the short-circuit "&&=", "||=" > and "//=". > > Taking the explanation in the section on "Assignment Operators" (further > below in perlop), the example given there for a "combined operator" can be > extended to a sequence of operators with "&&=": > $y &&= $x += 2; is equivalent to $y = $y && ($x = $x + 2); > Given the short-circuit behavior of '&&', it is clear that the assignment > "($x = $x + 2)" in the "equivalent assignment" (right) is not executed if $y > == 0 (and correctly the "simple" assignment '$y = ' is right-associative). OK so far. > But looking to the "original assignment" (left), where $y only appears to > the left of the "&&=" operator, the (effective) associativity for this "&&=" > operator is "left". I don't understand this at all. If &&= and += were left associative, then $y &&= $x += 2 would be parsed as ($y &&= $x) += 2, which is clearly not the case. What do you mean by effective associativity? -- Lukas Mai <plokinom@gmail.com>Thread Previous | Thread Next