Ç - 字符数组和字符指针

问题描述

为什么我无法定义
数组

 的char ** PP = {123，456，789};
 

不过，我可以把它定义为一个char * []，并将其发送到将接受它作为一个char功能**

 的char * PP [] = {123，456，789};
有趣（PP）;无效的乐趣（字符** pointerToPointer）
{
    // +（** pointerToPointer）; //错误
    的printf（％S，* pointerToPointer）;
}
//输出::123
 

和我为什么不能增加

  ++（** pointerToPointer）;
 

解决方案

要回答第一个问题，如果我们用指针的一个深度的原则可能会更清晰。这code是非法出于同样的原因：

 为int * PTR = {1，2，3};
 

在C，一个支撑初始化列表是不是一个对象（尤其不是一个数组）。它只能被视为物品从其中一个对象时被初始化读取初始化的列表。

PTR 是一个对象，所以最多有一个初始化可以采取，而初始的预期形式是指针（其中 1 不是）。

事实上，这code是在C11明确非法6.7.9 / 11：

However, there is a gcc bug/feature where it permits excess initializers for a scalar and ignores them. Further, some compilers may "be helpful" and "only" issue a warning, and initialize ptr to point to address 1, wherever that might be.

"scalar" means an object that's not a struct or an array.

---

Since C99 you can write:

int *ptr = (int []){1, 2, 3};

which creates an array (using the same storage duration as ptr) and points ptr at its first element.

This array is mutable; for a non-mutable one use int const *ptr = (int const[]){1, 2, 3}; instead.

---

Replacing int by char *, we see that you could write:

char **p = (char *[]){ "123", "456", "789" };

in which case the pointers in the array are mutable, but the things they point to (i.e. the string literals) still aren't.

Note that you should always use char const * when dealing with string literals, because they are not mutable. The fact that string literals have type char [N] is a historical hangover from before const was added to C. So:

char const **pp = (char const *[]){ "123", "456", "789" };

or with non-mutable pointers to strings:

char const *const *pp = (char const *const []){ "123", "456", "789" };

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