为什么是new int（*）[3]错误？ [英] Why is new int (*)[3] an error?

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):

expr.new]

5.3.4 New [expr.new]

....

new-type-id ： / em>

new-type-id:
    type-specifier-seq new-declarator_opt

new-declarator：

     > ptr-operator new-declarator _opt

     / em>

new-declarator:
    ptr-operator new-declarator_opt
    noptr-new-declarator

noptr-new-declarator：

     [  em>表达式 ]
attribute-specifier-seq _opt

     noptr -new-declarator  
[  constant-expression  ] 
attribute-specifier-seq
_opt

noptr-new-declarator:
    [ expression ]  attribute-specifier-seq_opt
    noptr-new-declarator  [ constant-expression ]  attribute-specifier-seq _opt

....

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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