What is wrong with print (2 & 2) >> 1?

问题

I am just wondering what happens with that piece of code. Why the result is incorrect only when printed directly, why is the newline ignored?

user@host_09:22 AM: perl
print 2 >> 1, "\n";
print 2 & 2, "\n";
print (2 & 2) >> 1, "\n";
1
2
2user@host_09:22 AM: perl
$a = (2 & 2) >> 1;
print "$a\n";
1

回答1:

When you print it with warnings it becomes clear(er)

perl -we'print (2 & 2), "\n"'

says

print (...) interpreted as function at -e line 1.
Useless use of a constant ("\n") in void context at -e line 1.

It works out print (2&2) as a function call to print^† and duly prints 2 (no newline!), and then it keeps evaluating the comma operator, with "\n" in void context next, which it also warns us about.

With >> 1 also there, the return 1 of print (2&2) (for success) is bit shifted to 0, which disappears into the void, and we get another "Useless use of ... in void context."

One fix is to add a + since what follows it must be an expression

perl -we'print +(2 & 2) >> 1, "\n"'

Or, make a proper call to print, with parenthesis around the whole thing

perl -we'print((2 & 2) >> 1, "\n")'

Both print a line with 1.

This is mentioned in print, and more fully documented in Terms and List operators and in Symbolic Unary operators, both in perlop.

---

^† It also warns about it as it is likely an error -- with a space before parens; no space, no warning.

回答2:

Perl interprets the parentheses as function arguments marker, as you can verify with

perl -MO=Deparse,-p -e 'print (2 & 2) >> 1'

Output:

(print(2) >> 1);

The canonical way is to precede the left parenthesis with a +:

print +(2 & 2) >> 1

来源：https://stackoverflow.com/questions/56556804/what-is-wrong-with-print-2-2-1