C 链表定义如何重新实现列表
嗨，这可能是一个愚蠢的问题，需要一个简单的解决方案，但我只是在互联网上找不到答案。
所以我正在为考试做练习并做作业。该程序的工作是找出链表中心的值是什么（如果列表的长度是奇数）
structdef 是：
typedef struct IntList IntList;
struct IntList {
int value;
IntList* next;
};
而我现在的确切问题是，当我尝试使用：
list = list->next;
我想在循环中逐步转到链表第 n 个位置（中心）的愿望列表。
有人知道我必须如何重写这个吗？如果您需要更多信息来帮助，请直接说出来，我会解释更多。
使用该函数，我检查列表的长度，在其他函数中，我有一个循环，仅到达长度的中间。
int length_list(IntList* list) {
int n = 0;
for(IntList* node = list; node != NULL; node = node->next) n++;
return n;
}
Hi this is probably a stupid question to ask with a simple solution but I just can't find an answer in the internet.
So I was exercising for an exam and worked on an assignment. The program has the job to find out what the value in the center of a linked list is (if the length of the list is an odd number)
The structdef is:
typedef struct IntList IntList;
struct IntList {
int value;
IntList* next;
};
and my exact problem right now is that I get a segmentation fault when I try using:
list = list->next;
I want to go step by step in a loop to go to the wished list at the nth position (the center) of the linked list.
Someone knows how I have to rewrite this? If you need more Information to help just say so and I will explain more.
With that function I check the length of the list and in my other function I have a loop which only goes to the mid of the length.
int length_list(IntList* list) {
int n = 0;
for(IntList* node = list; node != NULL; node = node->next) n++;
return n;
}
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循环结束后
for(IntList* node = list; node != NULL; node = node->next) n++;你肯定有node==NULL。< br>这并不是立即的问题。
但根据您对返回的
n值的处理方式，您可能会遇到相差一的问题。例如，在只有一个条目的列表中（毕竟 1 是奇数），尝试使用 1 太高的值可能会导致尝试访问不存在的节点。因此，我怀疑您的问题可以通过将循环更改为来解决
for(IntList* node = list; node->next != NULL; node = node->next) n++;，这样它就在最后一个现有节点上结束，而不是在后面。返回值会更低，无论你做什么都会“更加小心”。或者尝试与您显示并询问的小代码片段类似的操作，
list = list->next;仅当next不为 NULL 时才执行此操作，而不是 if只有list不为 NULL。After this loop ends
for(IntList* node = list; node != NULL; node = node->next) n++;you surely havenode==NULL.That is not immediatly a problem.
But depending on what you do with the value of
nwhich you return you might have an off-by-one problem. E.g. in a list with exactly one entry (1 is odd after all), the attempt to use a value which is 1 too high could result in an attempt to access a non-existing node.Because of this I suspect that your problem might be solved by changing the loop to
for(IntList* node = list; node->next != NULL; node = node->next) n++;, so that it ends on the last existing node, instead of behind. The return value will be lower, whatever you do with it will be "more-careful".That or try something similar with the small code fragment you show and ask about,
list = list->next;only do that if thenextis not NULL, not if onlylistis not NULL.