嵌套结构的指针

Question

这是我的代码......

#include <stdio.h>

struct one
{
    struct two
    {
            int r;
    }*b;
}*a;

void main()
{
    //struct two *new = &(*a).b;
    //new->r = 10;
    //printf("Value: %d", new->r);
    a = malloc(sizeof(struct one));
    //b = malloc(sizeof(struct two));
    (a->b)->r = 10;
    printf("Value: %d", (a->b)->r);
    return 0;

}

我在这里尝试的是,将结构定义为结构.现在两个对象都应该是指针.我想设置值,r然后显示它.

我得到它的唯一的事情Segmentation Fault 用gdb我得到了下面的,这似乎没有帮助太多..

(gdb) run
Starting program: /home/sujal.p/structtest/main

Program received signal SIGSEGV, Segmentation fault.
0x08048435 in main ()

我想知道如何执行上面提到的操作以及为什么这个东西会出现Segmentation错误.我已经尝试了一些网站上可用的方法,包括Stackoverflow的一些问题.

评论的行是我试图实现目标的失败,但失败了同样的错误.

编辑后尝试下面提到的技术..

void main()
{
    //struct two *new = &(*a).b;
    //new->r = 10;
    //printf("Value: %d", new->r);

    //a = malloc(sizeof(struct one));
    //a my_a = malloc(sizeof*my_a);
    //my_a->b = malloc(sizeof *my_a->b);
    //my_a->b->r = 10;
    //b = malloc(sizeof(struct two));
    //(a->b)->r = 10;
    //printf("Value: %d", my_a->b->r);

    a = (one*)malloc(sizeof(struct one));
    a->b = (one::two*)malloc(sizeof(struct one::two));
    (a->b)->r = 10;
    printf("Value: %d", (a->b)->r);
    return 0;

}

我已经尝试了所有提到的技术,他们给了我错误..我得到的最后一个错误如下..

new.c: In function âmainâ:
new.c:24:7: error: âoneâ undeclared (first use in this function)
new.c:24:7: note: each undeclared identifier is reported only once for each function it     appears in
new.c:24:11: error: expected expression before â)â token
new.c:25:13: error: expected â)â before â:â token
new.c:25:20: error: expected â;â before âmallocâ
new.c:28:2: warning: âreturnâ with a value, in function returning void [enabled by default]

Answer 1

您正在取消引用未初始化的指针.

您需要先分配一个实例struct one:

a = malloc(sizeof *a);

然后你可以初始化成员b:

a->b = malloc(sizeof *a->b);

然后你可以访问r:

a->b->r = 10;

这是一个有效的解决方案,通过调整代码和我的答案.