指針嵌套結構

Question

這裏是我的代碼...

#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; 

} 

我試圖在這裏，定義一個結構成的結構。現在這兩個對象都應該是指針。我想要設置r的值然後顯示它。

的唯一的事情我使用gdb我得到了下面的，這似乎沒有多大幫助得到它Segmentation Fault ..

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

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

我想知道如何執行提及的行動，爲什麼這件事情得到分割故障。我已經嘗試了一些網站上可用的方式，包括Stackoverflow的一些問題。

註釋行是我嘗試實現目標失敗，但失敗的同一個錯誤。

編輯試圖下文提到的技術後..

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; 

} 

我已經嘗試了所有的提到的技術，他們給我的錯誤..最後的錯誤我得到的是如下..

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

你去引用一個未初始化的指針。

您需要首先分配的struct one一個實例：

a = malloc(sizeof *a); 

，那麼你可以初始化成員b：

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

，然後你可以訪問r：

a->b->r = 10; 

Here is a working solution, by adapting your code with my answer。

Answer 2

你得到一個SIGSEGV因爲指針的間接引用謝勝利

(a->b)->r 

不是INI tialized /定義

來解決這個問題，你需要做以下

struct two 
{ 
     int r; 
} 

struct one 
{ 
two *b; 
}; 

one *a; 

... 

a = malloc(sizeof(struct one)); 
a->b = malloc(sizeof(struct two)); 

a->b->r = 10; 
printf("Value: %d", a->b->r); 
return 0; 

Answer 3

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

Answer 4

添加到@Quonux答案，我會以及定義類型：

typedef struct 
{ 
    int r; 
} t_two; 

typedef struct 
{ 
    t_two *p_b; 
} t_one; 

void main() 
{ 
    t_one *p_a; 

    p_a = malloc(sizeof(t_one); 
    p_a->p_b = malloc(sizeof(t_two)); 

    p-a->p_b->r = 10; 

    printf("Value: %d", p_a->p_b->r); 
    return 0; 

}