print的首字为括号时为何忽略掉下面的内容?

根本没提到 list context 字样。

QUOTE:
3.1. Terms and List Operators (Leftward)
Any term is of highest precedence in Perl. Terms include variables, quote and quotelike operators, most expressions in parentheses, or brackets or braces, and any function whose arguments are parenthesized. Actually, there aren't really any functions in this sense, just list operators and unary operators behaving as functions because you put parentheses around their arguments. Nevertheless, the name of Chapter 29 is Functions.

Now listen carefully. Here are a couple of rules that are very important and simplify things greatly, but may occasionally produce counterintuitive results for the unwary. If any list operator (such as print) or any named unary operator (such as chdir) is followed by a left parenthesis as the next token (ignoring whitespace), the operator and its parenthesized arguments are given highest precedence, as if it were a normal function call. The rule is this: If it looks like a function call, it is a function call. You can make it look like a nonfunction by prefixing the parentheses with a unary plus, which does absolutely nothing, semantically speaking--it doesn't even coerce the argument to be numeric.

For example, since || has lower precedence than chdir, we get:

chdir $foo    || die;       # (chdir $foo) || die
chdir($foo)   || die;       # (chdir $foo) || die
chdir ($foo)  || die;       # (chdir $foo) || die
chdir +($foo) || die;       # (chdir $foo) || die
but, because * has higher precedence than chdir, we get:
chdir $foo * 20;            # chdir ($foo * 20)
chdir($foo) * 20;           # (chdir $foo) * 20
chdir ($foo) * 20;          # (chdir $foo) * 20
chdir +($foo) * 20;         # chdir ($foo * 20)
Likewise for any numeric operator that happens to be a named unary operator, such as rand:
rand 10 * 20;               # rand (10 * 20)
rand(10) * 20;              # (rand 10) * 20
rand (10) * 20;             # (rand 10) * 20
rand +(10) * 20;            # rand (10 * 20)
In the absence of parentheses, the precedence of list operators such as print, sort, or chmod is either very high or very low depending on whether you look at the left side or the right side of the operator. (That's what the "Leftward" is doing in the title of this section.) For example, in:
@ary = (1, 3, sort 4, 2);
print @ary;         # prints 1324
the commas on the right of the sort are evaluated before the sort, but the commas on the left are evaluated after. In other words, a list operator tends to gobble up all the arguments that follow it, and then act like a simple term with regard to the preceding expression. You still have to be careful with parentheses:
# These evaluate exit before doing the print:
print($foo, exit);  # Obviously not what you want.
print $foo, exit;   # Nor this.

# These do the print before evaluating exit:
(print $foo), exit; # This is what you want.
print($foo), exit;  # Or this.
print ($foo), exit; # Or even this.
The easiest place to get burned is where you're using parentheses to group mathematical arguments, and you forget that parentheses are also used to group function arguments:
print ($foo & 255) + 1, "\n";   # prints ($foo & 255)
That probably doesn't do what you expect at first glance. Fortunately, mistakes of this nature generally produce warnings like "Useless use of addition in a void context" when warnings are enabled.

Also parsed as terms are the do {} and eval {} constructs, as well as subroutine and method calls, the anonymous array and hash composers [] and {}, and the anonymous subroutine composer sub {}.

Perl 中的圆括号有三种用途:提高优先级、暗示列表上下文、引起函数的参数。
本案属于第三种情况的用法。
另外,Perl 中还有一个专门的运算符:单目 +,用来解决此问题。
如果 print 后面的括号不希望成为 print 的参数,那么加个 + 号就没事了。
喏,就像这样:
print +($y*3)."\t".($y*3+1)."\t".($y*3+2)."\n";
再打个广告:
Perl 的单目 + 运算符,
作用挺像 Haskell 的 $ 运算符的。但是又不太一样,因为 $ 是双目运算符。
和上下文没关系的