java – 运算符优先级或最大蒙克规则首先出现在一元运算符中

在这里,我有以下代码：

int a,b,x;
a=b=1;
x=a+++b;

现在x的值将为2,因为a首先是后递增,然后将其添加到b.

以下是编译的字节代码：

 0  iconst_1
 1  dup
 2  istore_2 [b]
 3  istore_1 [a]
 4  iload_1 [a]
 5  iinc 1 1 [a]
 8  iload_2 [b]
 9  iadd
10  istore_3 [x]

所以表达式将等于x =(a)b.

现在另一个表达式x = a b,由于最大的munch规则而不会编译.它将变为x =(a)b,因此编译错误.

上述行为是x = a b,因为运算符的优先级还是因为最大的munch规则？

解决方法:

引自Lexical Translations：

The longest possible translation is used at each step, even if the
result does not ultimately make a correct program while another
lexical translation would.

Thus, the input characters a–b are tokenized (07001) as a, —, b,
which is not part of any grammatically correct program, even though
the tokenization a, –, –, b could be part of a grammatically correct
program.

这可以解释原因

x=a+++b

被解析为

x=(a++)+b

另一方面,b被标记为a,b,这导致错误.

标签：java,operator-precedence,unary-operator
来源： https://codeday.me/bug/20190612/1226204.html