在c ++中遇到奇怪的类型 [英] encountered strange type in c++

问题描述

这是什么意思？

What does this mean?

char const *  pWhiteList[] =
{
    "str1",
    "str2",
};

我知道 char * 意味着什么。



我也知道 char someVal [50] 意味着什么。



但是我从未见过如上所述的定义。



欢迎任何解释。



PS。我认为类似的构造在C ++中主要遇到，例如，

I know what char * means.

I also know what char someVal[50] means.

But I have never seen them defined together, like above.

Any explanations welcome.

PS. I think similar construct one encounters in main in C++, e.g.,

int _tmain(int argc, _TCHAR* argv[])

上面我们有 char * 和 [] 的组合：

_TCHAR* argv[]

。

这是什么意思？

.
So what does this mean?

推荐答案

char const *  pWhiteList[] =

从右到左阅读：

Reading right to left:

pWhiteList[] : is an array
*            : of pointers
const        : to constant
char         : characters

它声明一个常量指针到字符的数组并初始化它。

In换句话说，它会创建一个包含两个字符串的数组，并将其分配给pWhiteList。

It declares an array of constant pointer-to-characters and initialises it.
In other words it creates an array of two strings, and assigns it to pWhiteList.

读取C / C ++声明可能很难。

我也遇到了麻烦，直到最近我学会了可靠地阅读这些。



从实体名称开始（例如 x ）并执行以下操作：

Reading C/C++ declaration may be hard.
I had my troubles too until lately I learned to read these reliably.

Start at the entity name (e.g. x) and do the following:

1. 声明x为
2. 直接跟随（...）和 [...] 说函数（...）分别返回和数组[...]
3. 所有直接前缀指针运算符（例如 * ）说指向（添加const，volatile和其他指针运算符）
4. 如果以上嵌套（例如（* x））继续2。
5. 剩下的是类型，所以你说type

1. "declare x as "
2. for all the directly following (...) and [...] say "function (...) returning " and "array [...] of " respectively
3. for all directly prefixing pointer operators (e.g. *) say "pointer to " (add const, volatile and other pointer operators)
4. if the above was nested (e.g. (*x)) continue a 2.
5. what remains is the type, so you say "type"

例如

int x	声明x为 int
int * x	将x声明为 指向 int
const char * x [4]	声明x为 数组[4]的 指针to const char
const char（* x）[4]	声明x为 指向
long * f（double）	声明x为 函数（double）返回 指向 long
long（* f）（double）	声明x为 指向 函数（double）返回 long

干杯

Andi

E.g.

int x	declare x as int
int *x	declare x as pointer to int
const char *x[4]	declare x as array [4] of pointer to const char
const char (*x)[4]	declare x as pointer to array [4] of const char
long *f(double)	declare x as function (double) returning pointer to long
long (*f)(double)	declare x as pointer to function (double) returning long

Cheers
Andi

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