指向结构的指针在传递给函数时根本不改变 [英] Pointer to structure NOT changing at all when passed to function

问题描述

 #define MAX_STACK 10
 typedef int STACK_ITEM;
 typedef struct Stack *STACK ;

stack.c

 #include"stack.h"

struct Stack{
    STACK_ITEM contents[MAX_STACK];
    int tos;
};

_Bool create_stack(STACK s){
    s = malloc(sizeof(struct Stack));
    if(s == NULL )
        return 0;
    s->tos = 0;
    return 1;
}

当调用 create_stack 函数时，它根本不会影响 s(指向结构的指针).所以，我的问题是:为什么 s 不 改变，即使它是一个指针，而不是一个值，传递?

When calling the create_stack function, it doesn't affect s (pointer to structure) at all. So, my question is: why is s not changing, even though it is a pointer, not a value, passed?

推荐答案

记住，所有参数在C语言中都是按值传递的.

Remember, all parameters are passed by value in C.

当您将指针作为函数参数传递时，您可以访问该指针所指向的对象(或对象数组).

When you pass a pointer as a function parameter, you can access the object (or array of objects) pointed to by that pointer.

您的create_stack() 被传递一个指向struct Stack 的指针，即s 参数.然后它忽略 s 的原始值并重新分配它.这是允许的，但请注意它不会更改 create_stack() 调用者中的原始指针，因为函数参数是按值传递的.

Your create_stack() is passed a pointer to a struct Stack, that is the s parameter. It then ignores the original value of s and reassigns it. That is allowed, but note that it does not change the original pointer in the caller of create_stack() because function parameters are passed by value.

有两种方法可以做你想做的事.

There are two ways to do what you want.

第一种方式是将指针传递给指针:

The first way is to pass a pointer to a pointer:

_Bool create_stack(STACK *s){
    *s = malloc(sizeof(struct Stack));
    if(*s == NULL )
        return 0;
    (*s)->tos = 0;
    return 1;
}

称它为例如:

STACK mystack;
_Bool ret;

ret = create_stack(&mystack);
if (!ret) {
    /* error */
}

第二种方式是不传递任何东西，返回一个指向已分配堆栈的指针:

The second way is to not pass anything and return a pointer to the allocated stack:

STACK create_stack(void){
    STACK s = malloc(sizeof(struct Stack));
    if(s != NULL ) {
        s->tos = 0;
    }
    return s;
}

称它为例如:

STACK s;

s = create_stack();
if (s == NULL) {
    /* error */
}

最后，作为编程风格的一点，不要像你所做的那样typedef指针.如果指针是显式的，则更清楚，例如，使用 typedef struct Stack STACK; 而不是 typedef struct Stack *STACK; 并相应调整用法，例如，替换STACK s 替换为 STACK *s 并将 STACK *s 替换为 STACK **s.

Finally, as a point of programming style, do not typedef pointers in the way you have done. It is clearer if the pointer is explicit, for example, use typedef struct Stack STACK; instead of typedef struct Stack *STACK; and adjust the usage accordingly, for example, replacing STACK s with STACK *s and replacing STACK *s with STACK **s.

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