何时在C

Question

我正在遵循this C教程，并且正在上关于链表的课程，但是我有些困惑。第一个代码块具有以下内容：

#include <stdlib.h>

struct node {
  int x;
  struct node *next;
};

int main()
{
    /* This will be the unchanging first node */
    struct node *root;      

    /* Now root points to a node struct */
    root = (struct node *) malloc( sizeof(struct node) ); 

    /* The node root points to has its next pointer equal to a null pointer 
       set */
    root->next = 0;  
    /* By using the -> operator, you can modify what the node,
       a pointer, (root in this case) points to. */
    root->x = 5;     
 }

让我感到困惑的那一行是root = (struct node *) malloc( sizeof(struct node) );。因为在下面，所以还有另一个代码块：

#include <stdio.h>
#include <stdlib.h>

struct node {
  int x;
  struct node *next;
};

int main()
{
    /* This won't change, or we would lose the list in memory */
    struct node *root;       
    /* This will point to each node as it traverses the list */
    struct node *conductor;  

    root = malloc( sizeof(struct node) );  
    root->next = 0;   
    root->x = 12;
    /* rest of main */
    return 0;
}

在此块中，他们有root = malloc( sizeof(struct node ) );！为什么他们在最上面的块中键入强制转换malloc而不是第二个？有区别吗？