C 定义中的运算符优先级

发布于 2024-10-13 18:10:08 字数 284 浏览 5 评论 0原文

Wikipedia 声称 [] 运算符优先于求值中的 * 运算符。

那么，为什么下面的语句：

char *a[3];

声明一个 3 个字符指针的数组，而不是根据运算符优先级声明一个指向 3 个字符数组的指针？

原文

Wikipedia claims that the [] operator precedes the * operator in evaluation.

Then, why does the following statement:

char *a[3];

declare an array of 3 character pointers, rather than a pointer to an array of 3 characters as per the operator precedence?

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居里长安 2024-10-20 18:10:08

因为，正如维基百科所说，[] 的优先级高于 *？

处理声明时，a[3] 将在处理 * 之前作为“3 的数组”进行处理。

要声明指向三个字符数组的指针，必须使用括号来覆盖默认优先级：

char (*a)[3];

现在括号优先于数组。

Because, as Wikipedia says, [] has higher precedence than *?

Processing the declaration, the a[3] is processed as 'array of 3' before you process the *.

To declare a pointer to an array of three characters, you have to use parentheses to override the default precedence:

char (*a)[3];

Now the parentheses take precedence over the array.

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灰色世界里的红玫瑰 2024-10-20 18:10:08

以下是声明符的语法，取自标准 (§ 6.7.5)：

declarator:
    pointer_opt direct-declarator

direct-declarator:
    identifier
    ( declarator )
    direct-declarator [ type-qualifier-list_opt assignment-expression_opt ]
    direct-declarator [ static type-qualifier-list_opt assignment-expression ]
    direct-declarator [ type-qualifier-list static assignment-expression ]
    direct-declarator [ type-qualifier-list_opt * ]
    direct-declarator ( parameter-type-list )
    direct-declarator ( identifier-list_opt )

pointer:
    * type-qualifier-list_opt
    * type-qualifier-list_opt pointer

type-qualifier-list:
    type-qualifier
    type-qualifier-list type-qualifier

parameter-type-list:
    parameter-list
    parameter-list , ...

parameter-list:
    parameter-declaration
    parameter-list , parameter-declaration

parameter-declaration:
    declaration-specifiers declarator
    declaration-specifiers abstract-declarator_opt

identifier-list:
    identifier
    identifier-list , identifier

如您所见，[] 和 () 都绑定到 * 之前的声明符。取声明声明

int *a[N];

符为*a[N]，它符合上面的pointer_opt直接声明符模式，因此被解析为< code>*(a[N])，所以a是一个N元素的指针数组。

总结：

T *a[N]      -- declares an N-element array of pointer to T
T (*a)[N]    -- declares a pointer to an N-element array of T
T *f()       -- declares a function returning pointer to T
T (*f)()     -- declares a pointer to a function returning T

Here's the grammar for a declarator as taken from the standard (§ 6.7.5):

declarator:
    pointer_opt direct-declarator

direct-declarator:
    identifier
    ( declarator )
    direct-declarator [ type-qualifier-list_opt assignment-expression_opt ]
    direct-declarator [ static type-qualifier-list_opt assignment-expression ]
    direct-declarator [ type-qualifier-list static assignment-expression ]
    direct-declarator [ type-qualifier-list_opt * ]
    direct-declarator ( parameter-type-list )
    direct-declarator ( identifier-list_opt )

pointer:
    * type-qualifier-list_opt
    * type-qualifier-list_opt pointer

type-qualifier-list:
    type-qualifier
    type-qualifier-list type-qualifier

parameter-type-list:
    parameter-list
    parameter-list , ...

parameter-list:
    parameter-declaration
    parameter-list , parameter-declaration

parameter-declaration:
    declaration-specifiers declarator
    declaration-specifiers abstract-declarator_opt

identifier-list:
    identifier
    identifier-list , identifier

As you can see, both [] and () bind to the declarator before *. Take the declaration

int *a[N];

The declarator is *a[N], which fits the pointer_opt direct-declarator pattern above, and is thus parsed as *(a[N]), so a is an N-element array of pointer.

To sum up:

T *a[N]      -- declares an N-element array of pointer to T
T (*a)[N]    -- declares a pointer to an N-element array of T
T *f()       -- declares a function returning pointer to T
T (*f)()     -- declares a pointer to a function returning T

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涙—继续流 2024-10-20 18:10:08

我对这个问题感到困惑 - 声明的解释与运算符优先级相匹配。如果您想要一个指向数组的指针，则必须在 [] 绑定之前使用括号“将 * 绑定到标识符”。

char (*a)[3];

I'm confused about the question - the interpretation of the declaration matches the operator precedence. If you want a pointer to an an array you have to use parens to 'bind the * to the indentifier' before the [] binding.