c++ - typedef struct elt *堆栈；为什么这里有一个*

Question

这是使用链表实现 Stack 的完整代码。它来自 James Aspnes 为耶鲁大学编写的数据结构笔记(有什么好处吗？)

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

struct elt {
    struct elt *next;
    int value;
};

/* 
 * We could make a struct for this,
 * but it would have only one component,
 * so this is quicker.
 */
typedef struct elt *Stack;

#define STACK_EMPTY (0)

/* push a new value onto top of stack */
void
stackPush(Stack *s, int value)
{
    struct elt *e;

    e = malloc(sizeof(struct elt));
    assert(e);

    e->value = value;
    e->next = *s;
    *s = e;
}

int
stackEmpty(const Stack *s)
{
    return (*s == 0);
}

int
stackPop(Stack *s)
{
    int ret;
    struct elt *e;

    assert(!stackEmpty(s));

    ret = (*s)->value;

    /* patch out first element */
    e = *s;
    *s = e->next;

    free(e);

    return ret;
}

/* print contents of stack on a single line */
void
stackPrint(const Stack *s)
{
    struct elt *e;

    for(e = *s; e != 0; e = e->next) {
        printf("%d ", e->value);
    }

    putchar('\n');
}

int
main(int argc, char **argv)
{
    int i;
    Stack s;

    s = STACK_EMPTY;

    for(i = 0; i < 5; i++) {
        printf("push %d\n", i);
        stackPush(&s, i);
        stackPrint(&s);
    }

    while(!stackEmpty(&s)) {
        printf("pop gets %d\n", stackPop(&s));
        stackPrint(&s);
    }

    return 0;
}

我能看懂大部分代码。但我无法理解这部分typedef struct elt *Stack;

为什么Stack前面有一个*，它是什么意思？

我发现了指针的概念，尤其是函数的返回类型难以掌握。提前致谢。

Answer 1

I can understand most of the code. But I can't wrap my head around this part typedef struct elt *Stack;

那是使用 typedef*。这只是意味着当你写

Stack x;

在你的代码中它意味着:

struct elt * x;

如果你使用

Stack *s;

在你的代码中它意味着:

struct elt ** s;

I'm finding a concepts of pointers, especially in return types of functions hard to grasp. Thanks in advance.

然后我建议您在继续执行该代码之前先了解指针和指向指针的指针。

---

*_{我认为 C 和 C++ 中的 typedef 存在一些细微差别:see here}