placement of unary operator when using pointers in C

Question

I am trying to learn about pointers in C, and don't understand why the unary * operator was appended to the end of the word "node" in the following code snippet:

struct node* CopyList(struct node* head) {
    /* code here */
}

From my understanding of pointers, one can create a pointer with a statement like

int *pointerName;

and assign a "pointee" to it with a statement like

pointerName = malloc(sizeof(int));

and then dereference the pointer with a statement like

*pointerName = 4;

which will store the integer value 4 in the 4 bytes of memory (pointee location) which is "pointed to" by the pointerName pointer.

WITH THAT BEING SAID, what does it mean when the * is appended to the end of a word, as it is with

struct node*

???

Thanks in advance!

http://cslibrary.stanford.edu/103/

Answer 1

The location of the * ignores the whitespace between the base type and the variable name. That is:

int* foo; // foo is pointer-to-int
int *bar; // bar is also pointer-to-int

In both cases, the type of the variable is "pointer-to-int"; "pointer-to-int" is a valid type.

Armed with that information, you can see that struct node* is a type , that type being "pointer-to-node-structure". Finally, therefore, the whole line

struct node* CopyList(struct node* head)

means " CopyList is a function taking a pointer-to- struct node (called head ) and returning a pointer-to- struct node "

Answer 2

struct node* CopyList

To understand better you should read it from right to left. Which says CopyList is a function returning a pointer to node.

Answer 3

int *pointerName; is the same as int * pointerName; or int* pointerName; . The data type is int* in all those cases. So struct node* is just a pointer to struct node .

Answer 4

But it is suggested to use it with while declaring methods, like shown below 一起使用，如下所示

    struct node* CopyList(struct node* head) {
     /* code here */
    }

when declaring pointers of a type use * with the 使用* like shown below,

    int *ptr;

Declaring in that way increases readability.

For example consider this case,

    int* a,b,c;

The above statement is appearing like declaring three pointer variables of base type integer, actually we know that it's equals to

    int *a;
    int b,c;

Keeping the * operator near the data type is causing the confusion here, So following the other way increases readability, but it is not wrong to use * in either way.

Answer 5

node*表示以下函数/变量/结构的类型为“指向节点的指针”。