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 所有答案解释为什么你得到2而不是1实际上是错误的。 根据PHP文档，以这种方式混合+和++是未定义的行为，所以你可以获得1或2.切换到不同版本的PHP可能会改变你得到的结果，它将是一样有效。 
 
 见示例1 ，其中说： 
 
// mixing ++ and + produces undefined behavior
$a = 1;
echo ++$a + $a++; // may print 4 or 5
 
 笔记： 
 
 
 
 运算符优先级不能确定评估的顺序。 运算符优先级仅确定表达式$l + ++$l被解析为$l + (++$l) ，但不能确定是否首先计算+运算符的左或右操作数。 如果首先评估左操作数，结果将为0 + 1，如果首先评估右操作数，结果将为1 + 1。 
 
 
 操作员关联性也不影响评估顺序。 +运算符离开关联性仅确定$a+$b+$c被评估为($a+$b)+$c 。 它不会以什么顺序确定单个操作符的操作数被评估。 
 
 
 另外相关：在关于另一个表达式的错误报告中 ，一个PHP开发人员说：“我们不保证评估顺序[...]，就像C不一样，你可以指向在哪里说明首先对第一个操作数进行评估？ 

All the answers explaining why you get 2 and not 1 are actually wrong. According to the PHP documentation, mixing + and ++ in this manner is undefined behavior, so you could get either 1 or 2. Switching to a different version of PHP may change the result you get, and it would be just as valid.
 
See example 1, which says: 
 
// mixing ++ and + produces undefined behavior
$a = 1;
echo ++$a + $a++; // may print 4 or 5
 
Notes:
 
 
 
Operator precedence does not determine the order of evaluation. Operator precedence only determines that the expression $l + ++$l is parsed as $l + (++$l), but doesn't determine if the left or right operand of the + operator is evaluated first. If the left operand is evaluated first, the result would be 0+1, and if the right operand is evaluated first, the result would be 1+1.
 
 
Operator associativity also does not determine order of evaluation. That the + operator has left associativity only determines that $a+$b+$c is evaluated as ($a+$b)+$c. It does not determine in what order a single operator's operands are evaluated.
 
 
Also relevant: On this bug report regarding another expression with undefined results, a PHP developer says: "We make no guarantee about the order of evaluation [...], just as C doesn't. Can you point to any place on the documentation where it's stated that the first operand is evaluated first?"