FAQ 7.7: Why do Perl operators have different precedence than C operators?

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7.7: Why do Perl operators have different precedence than C operators?

    Actually, they don't. All C operators that Perl copies have the same
    precedence in Perl as they do in C. The problem is with operators that C
    doesn't have, especially functions that give a list context to
    everything on their right, eg. print, chmod, exec, and so on. Such
    functions are called "list operators" and appear as such in the
    precedence table in perlop.

    A common mistake is to write:

        unlink $file || die "snafu";

    This gets interpreted as:

        unlink ($file || die "snafu");

    To avoid this problem, either put in extra parentheses or use the super
    low precedence "or" operator:

        (unlink $file) || die "snafu";
        unlink $file or die "snafu";

    The "English" operators ("and", "or", "xor", and "not") deliberately
    have precedence lower than that of list operators for just such
    situations as the one above.

    Another operator with surprising precedence is exponentiation. It binds
    more tightly even than unary minus, making "-2**2" product a negative
    not a positive four. It is also right-associating, meaning that
    "2**3**2" is two raised to the ninth power, not eight squared.

    Although it has the same precedence as in C, Perl's "?:" operator
    produces an lvalue. This assigns $x to either $a or $b, depending on the
    trueness of $maybe:

        ($maybe ? $a : $b) = $x;


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