我在一个操作系统类中,我必须编写一个简单的堆栈程序(主要功能只是确定用户要求您做什么)。如果这不需要在 C 中,我很久以前就会这样做,但是因为我不太擅长 C 编码,所以它有一个“错误”......到目前为止的错误是它只是继续“弹出”相同的值关闭。 (它实际上并没有弹出任何东西)。我认为这是因为我不明白结构和指针是如何真正工作的。或者这是一个不那么明显的编码错误?
#include <stdio.h>
struct node {
int data;
struct node *next;
struct node *prev;
} first;
void push(int);
void pop();
int main(void)
{
int command = 0;
while (command != 3)
{
printf("Enter your choice:\n1) Push integer\n2) Pop Integer\n3) Quit.\n");
scanf("%d",&command);
if (command == 1)
{
// push
int num;
scanf("%d",&num);
push(num);
}
else
{
if (command == 2)
{
pop();
}
else
{
if (command != 3)
{
printf("Command not understood.\n");
}
}
}
}
return 0;
}
void push (int x)
{
struct node newNode;
newNode.data = x;
newNode.prev = NULL;
newNode.next = &first;
first = newNode;
printf("%d was pushed onto the stack.\n", first.data);
}
void pop()
{
if (first.data == '\0')
{
printf("Error: Stack Empty.\n");
return;
}
printf("%d was popped off the stack.\n", first.data);
first = *(first.next);
first.prev = NULL;
}
最佳答案
首先应该是一个指针。将其更改为 struct node *first ;
在主要初始化 first=NULL ;
改变你推送/弹出操作如下,
void push (int x)
{
struct node *newNode;// It should be a pointer
newNode = (struct node *)malloc(sizeof(struct node));
newNode->data = x;
//newNode.prev = NULL; // You don't need this
newNode->next = first;
first = newNode;
printf("%d was pushed onto the stack.\n", first->data);
}
void pop()
{
struct node *prevPtr;
//if (first.data == '\0')
if (first == NULL) // check if stack is empty
{
printf("Error: Stack Empty.\n");
return;
}
printf("%d was popped off the stack.\n", first->data);
prevPtr = first;
first = first->next;
free(prevPtr);
}
关于c - C中的简单堆栈程序,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/18995071/