它看起来像一个printf使我的程序bug

Question

我正在学习结构和链表。不过，我面临的一个问题是，我无法调试我的程序的错误，因为它来自于函数printf，这是我用来调试程序的。

下面的程序工作正常：

struct pointer_struct 
{ 
    struct new_struct *ptr; 
}; 

struct new_struct 
{ 
    int i; 
    struct new_struct *ptr; 
}; 

void init(struct pointer_struct *pointer, int nb) 
{ 
    struct new_struct my_struct; 
    my_struct.i = nb; 
    my_struct.ptr = NULL; 
    pointer->ptr = &my_struct; 
} 

int main(void) 
{ 
    struct pointer_struct pointer; 
    pointer.ptr = NULL; 

    init(&pointer, 15); 
    //printf("pointer.ptr = %p\n", pointer.ptr); 
    printf("pointer.ptr->i = %d\n", pointer.ptr->i); 
} 

输出：

pointer.ptr->i = 15 

但只要我去掉了注释行，i需要怪异值。以下是一些输出示例：

$./a.out 
pointer.ptr = 0x7fffc6bcc650 
pointer.ptr->i = -448723664 
$./a.out 
pointer.ptr = 0x7fffd09ed480 
pointer.ptr->i = 1218512176 
$./a.out 
pointer.ptr = 0x7ffff630fa70 
pointer.ptr->i = -1073674960 

printf怎么回事？

Answer 1

你用一个局部变量初始化pointer.ptr。

void init(struct pointer_struct *pointer, int nb) 
{ 
    struct new_struct my_struct; 
    my_struct.i = nb; 
    my_struct.ptr = NULL; 
    pointer->ptr = &my_struct; // MISTAKE!!! my_struct is on the stack. 
           // its memory space could be overwritten at 
           // any time after the function returns. 
} 

后来，在主

printf("pointer.ptr = %p\n", pointer.ptr); // This call to printf uses the stack, 
              // and overwrites the space used 
              // by my_struct 
    printf("pointer.ptr->i = %d\n", pointer.ptr->i); 

Answer 2

你有一个未定义的行为或UB，它总是一个坏事™。

void init(struct pointer_struct *pointer, int nb) 
{ 
    struct new_struct my_struct; 
    my_struct.i = nb; 
    my_struct.ptr = NULL; 
    pointer->ptr = &my_struct; 
} // here my_struct lifetime is finish so pointer->ptr become invalid 

int main(void) 
{ 
    struct pointer_struct pointer; 
    pointer.ptr = NULL; 

    init(&pointer, 15); 
    printf("pointer.ptr = %p\n", pointer.ptr); // pointer.ptr is not valid so it's UB 
    printf("pointer.ptr->i = %d\n", pointer.ptr->i); // This is UB too 
}