鏈接列表推送功能

Question

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

struct stackNode 
{ 
    int data; 
    struct stackNode *nextPtr; 
}; 


void instructions() 
{ 
    printf("[1]Push a value on the stack\n"); 
    printf("[2]Pop a value off the stack\n"); 
    printf("[3]Display the whole stack\n"); 
    printf("[4]Exit"); 
} 

void push(struct stackPtr *topPtr, int info) 
{ 
    struct stackPtr *newPtr; 
    newPtr= malloc(sizeof(struct stackNode)); 
    if(newPtr !=NULL) 
    { 
     newPtr->data = info; 
     newPtr->nextPtr=*topPtr; 
     *topPtr=newPtr; 
    } 
    else 
    { 
     printf("%d not inserted no memory available"); 
    } 

int main() 
{ 
    struct StackNodePtr *stackPtr; 
    stackPtr = NULL; 
    int choice, value; 
    do 
    { 
     instructions(); 
     printf("\nEnter Your Choice: "); 
     scanf("%d",&choice); 
     if(choice == 1) 
     { 
      printf("Enter a value for the stack");  
     }    
     if(choice == 2) 
     { 
      printf(" "); 
     }  
     if(choice == 3) 
     { 
      printf(" "); 
     } 
     if(choice == 4) 
     { 
      printf("bye!"); 
      return 0; 
     } 
    } while(choice !=4); 
    system("pause"); 
} 

我做了一個函數推送我的鏈表和堆棧代碼，但事情是它不工作有很大的錯誤，在功能推它有什麼問題嗎？它不允許使用malloc爲什麼？

Answer 1

// just precede the struct type with ... struct 
struct stackNode* stackPtr = NULL; 

快樂編碼。

Answer 2

這是否適合您？

struct stackNode { int data; struct stackNode *nextPtr; }; 

int main() { struct stackNode * stackPtr = NULL; } 

Answer 3

struct stackNode 
{ 
    int data; 
    struct stackNode *nextPtr; 
}; 

int main() 
{ 
    struct stackNode *stackPtr = NULL; 
} 

Answer 4

這條線：

typedef struct StackNode *StackNodePtr; 

是錯誤的方式。你的類型定義應該是：

typedef struct stackNode StackNode; 
typedef StackNode* StackNodePtr; 

不知道爲什麼你反對使用typedef - 它往往使代碼更大量的可讀性。