int**foo[][]（）中的foo是什么-whatisfooinint**foo[][]()

1  

It is nothing specific. Meaning that on the surface it looks like a C declaration, but it is not well-formed. It is illegal and as such it means nothing.

这没什么特别的。这意味着它表面看起来像一个C声明,但它没有很好的形成。这是非法的,因此它没有任何意义。

Firstly, it appears like a two-dimensional array declaration, but in C language an array declaration is required to specify all sizes except possibly for the very first one. Your declaration omits the second size as well, which makes it illegal.

首先,它看起来像一个二维数组声明,但在C语言中,需要一个数组声明来指定除第一个之外的所有大小。您的声明也省略了第二个尺寸,这使其成为非法。

Secondly, even if we ignore the missing sizes, it looks like a declaration for an array of functions. It is illegal to declare arrays of functions in C.

其次,即使我们忽略了缺少的大小,它看起来像一个函数数组的声明。在C中声明函数数组是非法的。

For example, this would make a legal C declaration

例如,这将产生合法的C声明

int (**foo[][42])()

but not what you have originally.

但不是你原来的。

0  

The syntax of this declaration is that foo[][] declares a 2-D array (or it would, if the second bound had a dimension specified - as it stands that's illegal); and then the rest of it is:

这个声明的语法是foo [] []声明一个2-D数组(或者,如果第二个绑定指定了一个维度,它将会这样做 - 因为它表示这是非法的);然后剩下的就是:

int **bar();    // with bar = foo[][]

which is a function taking unspecified arguments and returning int **. However, since bar is an array type here this attempts to declare an array of functions, which is illegal. (Not to be confused with an array of function pointers, which would be OK).

这是一个接受未指定参数并返回int **的函数。但是,由于bar是一种数组类型,因此尝试声明一组函数,这是非法的。 (不要与函数指针数组混淆,这样就可以了)。

The grammar rules are that the ** bind to the int (not to the bar) unless you use parentheses to force them to bind to the bar; so they are part of the type int ** and they are not saying that bar is a pointer.

语法规则是**绑定到int(而不是bar),除非你使用括号来强制它们绑定到bar;所以他们是int **类型的一部分,他们并不是说bar是指针。