java - c/c++ 指针指向数组和数据类型判断时会不会有二义性？

Question

{代码...} 两个理解：1.rowlist是个动态数组指向一大块内存，数组里的元素存储着指向LinkList&lt;int&gt;链表的地址。即rowlist是指针数组。如图一 2.rowlist指向一个内存，这个内存存储着&lt;LinkList&lt; int &...

伊谢尔伦 · Answer

int dd=1;
int *a=ⅆ    // 1
int *c=new int[2];  //2

The subject said:

Here 1 indicates that a is a pointer of type int, and 2 indicates that c is a pointer of type int array. Right?

That’s not the case, a and c are actually the same thing, they are pointers to int. c is not a "pointer of type int array".

Why is c originally an int pointer, but it can actually point to an int array? Because when the int array is assigned to the int pointer, the C language secretly converts it (the int array) into the first element of the int array. The professional term for this "sneak conversion" is called "implicit conversion".

So to summarize:

The variable c is not a pointer of type int array, but a pointer of type int;
Then why variable c can point to an int array? It’s because there is an implicit conversion;

PHP中文网 · Answer

int *c=new int[2];  //2

c is a pointer to int. Pointer to array should be written as

int (*c)[2]=new int[2];

How many bytes will the pointer +1 move? Give it a try.