问题描述
下面的程序打印了相同的号码在GCC 4.8.2：
The following program prints the same number twice on gcc 4.8.2:
#include <stdio.h>
int main()
{
char a[13];
printf("sizeof a is %zu\n", sizeof a );
printf("sizeof(a) is %zu\n", sizeof(a));
}
据this reddit的帖子，海湾合作委员会是在这方面不符合标准的，因为一个括号前pression不例外的，当阵列到指针腐烂不会发生在名单上。
According to this reddit post, gcc is not standard-conformant in this respect, because a parenthesized expression is not on the list of exceptions for when array-to-pointer decay does not happen.
这家伙是正确的？下面是相关标准的报价：
Is this guy correct? Here is the relevant standard quote:
除非它是的sizeof 的运算数或一元＆安培; 运营商，或者是一个字符字符串用来初始化字符类型的数组，或者是用于初始化元素类型与 wchar_t的兼容的，即具有类型的阵列左值数组中的宽字符串文字类型'转换到具有键入前pression'指向类型为指向数组对象的初始成员，是不是左值。
只是要清楚，他认为，（一）应触发数组到指针的衰减，因为括号没有在上面（<$ C $名单所涉及C>的sizeof 符，一元＆放大器; 运营商，字符串作为初始化）
Just to be clear, he argues that (a) should trigger array-to-pointer decay, because parentheses are not covered in the list above (sizeof operator, unary & operator, string literal as initializer).
推荐答案
无论是看似多余的括号影响程序的语义是在C标准还没有得到充分解决了长期存在的问题。
Whether seemingly redundant parentheses affect the semantics of a program is a long-standing issue in the C standard that still hasn't been adequately resolved.
人们普遍声称，（（无效*）0）在技术上是不是一个空指针常数，因为没有规定说一个parenthesised空指针常量是一个空指针常量。
It is commonly claimed that ((void*)0) is technically not a null pointer constant, because there is no rule that says a parenthesised null pointer constant is a null pointer constant.
一些编译器问题为个char [] =（ABC）的错误; ，因为当一个字符数组可以从一个字符串初始化文字，该规则没有按T盖parenthesised字符串。
Some compilers issue an error for char s[] = ("abc");, because while a character array can be initialised from a string literal, that rule doesn't cover parenthesised string literals.
有很多类似的例子。你已经找到了其中的一个。
There are many similar examples. You've found one of them.
从我所知道的，concensus基本上是规则的应的是什么C ++做什么，但C从不正式通过。 C ++使得parenthesised前pression功能等效非parenthesised前pression，与几个明确态异常。这将包括所有这些问题的一次。
From what I can tell, the concensus is basically that the rule should be what C++ does, but what C never formally adopted. C++ makes a parenthesised expression functionally equivalent to the non-parenthesised expression, with a few explicitly-stated exceptions. This would cover all those issues at once.
所以从技术上来说，这个家伙可能被认为是正确的，但它是标准的过于严格的跨pretation，没有人真正遵循，因为它是人所共知的标准仅仅是故障在这里。
So technically, the guy could be considered correct, but it's an overly strict interpretation of the standard that nobody really follows, since it's common knowledge that the standard is simply faulty here.
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