# 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" produce 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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