数组声明方括号内的星号在C中意味着什么 [英] What does asterisk inside square bracket of array declaration mean in C

问题描述

我正在编写一个自定义C99解析器.我从这获得了语法链接.这个语法说以下是声明数组的有效语法-

I am writing a custom C99 parser. I got the grammar from this link. This grammar says following is a valid syntax for declaring arrays -

int arr[*];

语法的相关部分如下-

direct-declarator ::=
    identifier
    "(" declarator ")"
    direct-declarator "[" type-qualifier-list? assignment-expression? "]"
    direct-declarator "[" "static" type-qualifier-list? assignment-expression "]"
    direct-declarator "[" type-qualifier-list "static" assignment-expression "]"
    direct-declarator "[" type-qualifier-list? "*" "]"
    direct-declarator "(" parameter-type-list ")"
    direct-declarator "(" identifier-list? ")"

我尝试使用gcc使用此声明编译代码.它给了我以下警告-

I tried compiling a code with this declaration using gcc. It gave me following warning -

错误:函数原型范围以外的其他地方不允许使用"[*]"

error: ‘[*]’ not allowed in other than function prototype scope

因此，我尝试使用这种语法声明函数原型，并且在编译时没有任何错误或警告.我没有得到的是这种语法在语义上可能意味着什么.有解释的专家吗?

So I tried declaring a function prototype with this type of syntax and it compiled without any error or warning. What I am not getting is what can this syntax possibly mean semantically. Any expert with an explanation?

推荐答案

它是未指定大小的可变长度数组的声明器.另外，以下声明

It's a declarator for a variable length array with unspecified size. Furhtermore, the following declaration

void func(size_t n, char s[n]);

等同于简单地写:

void func(size_t n, char s[*]);

以上内容对于编写​​标头特别有用，在标头中，通常只声明参数类型

The above is particularly useful for writing headers, where you'd normally declare only the parameter types

void func(size_t, char [*]);

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