C 为什么动态分配的变量不会像静态变量那样改变(当我们使用它们的地址时)
head是指向内存的指针,它等于当前值C 为什么动态分配的变量不会像静态变量那样改变(当我们使用它们的地址时),c,dynamic,linked-list,static,static-variables,C,Dynamic,Linked List,Static,Static Variables,head是指向内存的指针,它等于当前值 void display(node_t **head) { struct node *current = (*head); if((*head) == NULL) { printf("List is empty \n"); return; } while(current != NULL) { printf("current->
void display(node_t **head) {
struct node *current = (*head);
if((*head) == NULL) {
printf("List is empty \n");
return;
}
while(current != NULL) {
printf("current->n=%d\n", current->next); //0 1 2
printf("head->n=%d\n", head->next); // 0 0 0 if we assign current to its next
current = current->next; // should it affect to the head because we are
//because we are assigning memory address to another
}
}
使用静态变量的示例代码:
int *head,*current,c=1;
head=&c;
current=&c;
(*current)++;
在这种情况下,head和current将发生变化,因为此时:
struct node*current=(*head)代码>取消对头部的引用。如果head是一个节点指针数组,那么现在current指向该数组的第一项
此外,我建议在解除对磁头的引用之前检查NULL
,否则将出现SEGFAULT
忽略不一致的命名(是n
还是next
?),head
和current
是独立的对象-您对current
所做的任何操作都不会影响head
如果我们画一幅画可能会有帮助。假设您有一个包含三个节点的列表(我不知道您的节点结构是什么样子,所以这是非常通用的):
您有一个指向第一个元素的指针(我们称之为h
):
display
函数中的head
参数指向h
:
+---+ +---+---+ +---+---+ +---+---+
h: | |----->| | |----->| | |------>| | |---|||
+---+ +---+---+ +---+---+ +---+---+
+---+
head: | |
+---+
|
V
+---+ +---+---+ +---+---+ +---+---+
h: | |----->| | |----->| | |------>| | |---|||
+---+ +---+---+ +---+---+ +---+---+
您使用*head
的值初始化current
,该值为h
,因此current
指向列表的第一个节点:
+---+
head: | |
+---+
|
V
+---+ +---+---+ +---+---+ +---+---+
h: | |----->| | |----->| | |------>| | |---|||
+---+ +---+---+ +---+---+ +---+---+
^
|
+---+
current: | |
+---+
+---+
head: | |
+---+
|
V
+---+ +---+---+ +---+---+ +---+---+
h: | |----->| | |----->| | |------>| | |---|||
+---+ +---+---+ +---+---+ +---+---+
^
|
+---+
current: | |
+---+
+---+
head: | |
+---+
|
V
+---+ +---+---+ +---+---+ +---+---+
h: | |----->| | |----->| | |------>| | |---|||
+---+ +---+---+ +---+---+ +---+---+
^
|
+---+
current: | |
+---+
每次你执行
current = current->next
这将设置current
以指向列表中的下一个元素:
+---+
head: | |
+---+
|
V
+---+ +---+---+ +---+---+ +---+---+
h: | |----->| | |----->| | |------>| | |---|||
+---+ +---+---+ +---+---+ +---+---+
^
|
+---+
current: | |
+---+
+---+
head: | |
+---+
|
V
+---+ +---+---+ +---+---+ +---+---+
h: | |----->| | |----->| | |------>| | |---|||
+---+ +---+---+ +---+---+ +---+---+
^
|
+---+
current: | |
+---+
+---+
head: | |
+---+
|
V
+---+ +---+---+ +---+---+ +---+---+
h: | |----->| | |----->| | |------>| | |---|||
+---+ +---+---+ +---+---+ +---+---+
^
|
+---+
current: | |
+---+
您对当前所做的任何操作都不会影响头部-它们是独立的对象
在这种情况下,水头和电流将发生变化
是的,因为您正在显式地将新值分配给head
和current
——但是,current++
对head
没有影响
顺便说一句,这不是有效的C,看起来你是从C开始,用Python结束的。请将两个版本的代码发布为,并显示两种情况下的预期输出与实际输出的比较head->n
无效,因为节点*
没有成员。current=current->next代码>不会影响head
,因为current
和head
是不同的变量。您的第二个代码(如果您对其进行了修复,使其编译)也不会更改head
int *head,*current,c=1;
head=&c;
current=&c;
while(True):
current++;