问题描述
typedef int (*A)[3];
int (**p)[3] = new A; // OK
int (**q)[3] = new (int(*)[3]); // OK
int (**r)[3] = new int (*)[3]; // error
来自GCC的错误是错误：expected primary-expression before '）'token 。
The error from GCC is error: expected primary-expression before ')' token . Why are the extra parentheses required in this expression?
推荐答案
标准定义了 new-type-id 作为新声明符的最长序列。还有一个注释，说明类似的情况（虽然它分配函数的指针）：
The standard defines new-type-id as the longest sequence of new-declarators. There is also a note, which illustrates a similar situation (although it allocates pointers to functions):
....
new-type-id ： / em>
new-type-id:
type-specifier-seq new-declarator
new-declarator：
> ptr-operator new-declarator
/ em>
new-declarator:
ptr-operator new-declarator
noptr-new-declarator
noptr-new-declarator：
[ em>表达式 ]
attribute-specifier-seq
noptr -new-declarator
[ constant-expression ]
attribute-specifier-seq
noptr-new-declarator:
[ expression ] attribute-specifier-seq
noptr-new-declarator [ constant-expression ] attribute-specifier-seq
....
新表达式中的 new-type-id 新声明符。
[注意：这可以防止声明符运算符& ，&& ， * 和 [] 和
。 - ：
The new-type-id in a new-expression is the longest possible sequence of new-declarators. [ Note: this prevents ambiguities between the declarator operators &, &&, *, and [] and their expression counterparts. — end note ] [ Example:
new int * i; // syntax error: parsed as (new int*) i, not as (new int)*i
* 是指针声明符，而不是乘法运算符。 - ]
The * is the pointer declarator and not the multiplication operator. — end example ]
/ em>的新表达式可以
具有惊人的效果。 [示例：
[ Note: parentheses in a new-type-id of a new-expression can have surprising effects. [ Example:
new int(*[10])(); // error
是错误的，因为绑定是
(new int) (*[10])(); // error
相反， new operator
可用于创建复合类型的对象（3.9.2）：
Instead, the explicitly parenthesized version of the new operator can be used to create objects of compound types (3.9.2):
new (int (*[10])());
为函数分配一个10个指针的数组（不带参数，
返回 int 。 - end example ] - 结束注释]
allocates an array of 10 pointers to functions (taking no argument and returning int. — end example ] — end note ]
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