指针之间的区别的指针和指针数组？ [英] Difference between pointer to pointer and pointer to array?

问题描述

由于数组的名称实际上是一个指向数组的指针，以下code中的第一个元素：

Given that the name of an array is actually a pointer to the first element of an array, the following code:

#include <stdio.h>

int main(void)
{
    int a[3] = {0, 1, 2};
    int *p;

    p = a;

    printf("%d\n", p[1]);

    return 0;
}

打印 1 ，符合市场预期。

现在，因为我可以创建一个指向指针的指针，我写了以下内容：

Now, given that I can create a pointer that points to a pointer, I wrote the following:

#include <stdio.h>                                                              

int main(void)                                                                  
{                                                                               
        int *p0;                                                                
        int **p1;                                                               
        int (*p2)[3];                                                           
        int a[3] = {0, 1, 2};                                                   

        p0 = a;                                                                 
        p1 = &a;                                                                
        p2 = &a;                                                                

        printf("p0[1] = %d\n(*p1)[1] = %d\n(*p2)[1] = %d\n",                    
                        p0[1], (*p1)[1], (*p2)[1]);                             

        return 0;                                                               
}

我预期编印

p0[1] = 1
(*p1)[1] = 1
(*p2)[1] = 1

但是，相反，它出错在编译的时候，他说：

But instead, it goes wrong at compile time, saying:

test.c: In function ‘main’:
test.c:11:5: warning: assignment from incompatible pointer type [enabled by default]

这是为什么分配错了吗？如果 P1 是一个指向一个指向 INT 和 A 是一个指向 INT （因为它是一个数组 INT S的名字），为什么不能' T I分配＆放大器;一个到 P1

Why is that assignment wrong? If p1 is a pointer to a pointer to an int and a is a pointer to an int (because it's the name of an array of ints), why can't I assign &a to p1?

推荐答案

11号线

        p1 = &a;

其中， P1 的类型为 INT ** 和 A 已键入 INT [3] ，对吧？

where p1 has type int ** and a has type int[3], right?

好; ＆放大器;一个已键入 INT（*）[3] 以及该类型是不兼容 INT ** 因为编译器告诉你。

Well; &a has type int(*)[3] and that type is not compatible with int** as the compiler told you

您可能想尝试

        p1 = &p0;

和阅读的C-FAQ ，特别是第6。

在短：数组不是指针和指针不是数组

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