不使用printf()时出现malloc错误

Question

#include <stdio.h>
#include <stdlib.h>
typedef struct lis
{
    int num;
    struct lis * next;
} list;
void  fun(list ** h, int nu) {
    *h = malloc(sizeof(list)*nu);
    list *p = *h;
    int i=1;
    list * nextx;
    while(i<=nu) {
        nextx = p + 1;
        p->num = i;
        p->next = nextx;
        //printf("%d\n", nextx);
        p += 1;
        i++;
    }
    p->next = NULL;
}


int main(int argc, char const *argv[])
{
    list * first = NULL;
    fun(&first,10);
    free(first);
    return 0;
}

我正在学习c中的列表

无论何时运行此代码,都会产生malloc错误

如果我注释掉printf("%d\n", nextx);哪个显示下一个节点就可以了.

怎么了？

Answer 1

在循环的最后一次运行中,您的代码执行:

nextx = p+1; // points one past the last array' element
p->num = nu-1; // ok
p->next = p+1; // probably not what you wanted, but not a fault per se
p += 1; // This is the cause of your problem
i++; // out of the loop...
p->next = NULL; // dereference out of array pointer!

退出循环前一步,然后正确设置最后一个元素:

while (i<nu) {
   ...
}
p->next = NULL;
p->num = nu-1;