它看起來像一個printf使我的程序bug

Question

我正在學習結構和鏈表。不過，我面臨的一個問題是，我無法調試我的程序的錯誤，因爲它來自於函數printf，這是我用來調試程序的。

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); 
} 

輸出：

pointer.ptr->i = 15 

但只要我去掉了註釋行，i需要怪異值。以下是一些輸出示例：

$./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怎麼回事？

Answer 1

你用一個局部變量初始化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. 
} 

後來，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

你有一個未定義的行爲或UB，它總是一個壞事™。

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 
}