c - 函数调用期间指针到指针不起作用？

Question

我正在尝试编写一个带有堆栈操作辅助函数的单独文件。我想通过引用将堆栈顶部作为参数传递给主文件中的堆栈操作。

由于 top 正在被修改，我将通过引用传递指针 top 。但即便如此，它也不起作用。我哪里出错了？

P.S.:我知道这不是实现 Stack 的最佳方式，但我只是想了解为什么它不起作用。

//堆栈.h

void print(stacknode **P)
{

    stacknode *S;
    S=*P;

    printf("Printing stack from top to bottom...\n");
    stacknode *temp=S;
    while(temp != NULL)
    {
        printf("%d\t", temp->data);
        temp=temp->next;
    }
    printf("\n");
}


void push(stacknode **P, int n)

{

    stacknode *S;
    S=*P;
    stacknode *new=(stacknode *)malloc(sizeof(stacknode));
    new->data=n;
    new->next=S; 
    S=new;
    print(&S);

}

//main.c

main()
{
    printf("Creating new stack...\n");
    stacknode *S=NULL;

    printf("Pushing first number....\n");
    push(&S, 2);

    print(&S);/*Prints nothing*/

}

Answer 1

Since top is getting modified, I am passing the pointer top by reference.

但是你不会用这个事实来改变顶部。这是一个解决方案(我尚未编译或测试它，因此它可能包含错误):

Stack.h:(仅在头文件中声明，无代码)

typedef struct stacknode stacknode;
struct stacknode {
    stacknode* next;
    int data;
};

void print(stacknode* top); // no need for ptr ref
void push(stacknode** ptop);

堆栈.c:

#include "Stack.h"
#include <stdio.h>

void print(stacknode* top)
{
    printf("Printing stack from top to bottom...\n");
    for (stacknode* p = top; p; p = p->next)
    {
        printf("%d\t", p->data);
    }
    printf("\n");
}

void push(stacknode** ptop, int n)
{
    stacknode* p = malloc(sizeof *p); // don't cast malloc in C
    if (!p)
        /* handle out of memory */;
    p->data = n;
    p->next = *ptop; 
    *ptop = p;
    print(p);
}

main.c:

#include "Stack.h"
#include <stdio.h>

int main(void) // declare return type
{
    printf("Creating new stack...\n");
    stacknode* S = NULL;

    printf("Pushing first number....\n");
    push(&S, 2);

    print(S);
    return 0;
}