perl - Perl 如何决定对表达式中的项求值的顺序？

Question

鉴于代码:

my $x = 1;

$x = $x * 5 * ($x += 5);

我希望 $x成为 180 :

$x = $x * 5 * ($x += 5); #$x = 1
$x = $x * 5 * 6;         #$x = 6
$x = 30 * 6;
$x = 180;
180;

但它是 30 ;但是，如果我更改条款的顺序:

$x = ($x += 5) * $x * 5;

我确实收到 180 .我感到困惑的原因是 perldoc perlop 非常直白地说:

A TERM has the highest precedence in Perl. They include variables, quote and quote-like operators, any expression in parentheses, and any function whose arguments are parenthesized.

自 ($x += 5)在括号中，它应该是一个术语，因此无论表达式的顺序如何，都首先执行。

Answer 1

输入问题的行为给了我答案:术语具有最高优先级。这意味着 $x在第一块代码中被评估并产生 1 ，然后 5被评估并产生 5 ，然后 ($x += 5)评估并产生 6 (具有将 $x 设置为 6 的副作用):

$x = $x * 5 * ($x += 5);
address of $x = $x * 5 * ($x += 5); #evaluate $x as an lvalue
address of $x = 1 * 5 * ($x += 5);  #evaluate $x as an rvalue
address of $x = 1 * 5 * ($x += 5);  #evaluate 5
address of $x = 1 * 5 * 6;          #evaluate ($x += 5), $x is now 6
address of $x = 1 * 5 * 6;          #evaluate 1 * 5
address of $x = 5 * 6;              #evaluate 1 * 5
address of $x = 30;                 #evaluate 5 * 6
30;                                 #evaluate address of $x = 30

同样，第二个例子像这样减少:

$x = ($x += 5) * $x * 5; 
address of $x = ($x += 5) * $x * 5; #evaluate $x as an lvalue
address of $x = 6 * $x * 5;         #evaluate ($x += 5), $x is now 6
address of $x = 6 * 6 * 5;          #evaluate $x as an rvalue
address of $x = 6 * 6 * 5;          #evaluate 5
address of $x = 36 * 5;             #evaluate 6 * 6
address of $x = 180;                #evaluate 36 * 5
180;                                #evaluate $x = 180