Question

我正在写这段代码：

#include<stdio.h>
#include<stdlib.h>
int main()
{
FILE *fp;
int i;
fp = fopen("keimeno.txt","r");
fscanf(fp,"%d",i);
printf("%d\n",i);
    fclose(fp);
return 0;
}

，文件包含：

2
Yiannis Ioannou 356
3
Today
10347
If
345
And then none
1542
John Smith 743
2
My story
3940
Feedback
682
END

当我尝试运行它时，它会退出值3221225477而不是打印数字2 ..

任何人都可以解释原因吗？

Answer 1

扫描数字时，需要传递要存储结果的变量的地址：

fscanf(fp,"%d",&i);

你有

的地方

fscanf(fp,"%d",i);
               ^  missing the & sign!

您的编译器真的应该警告过您 - 编译时是否启用了警告？

这里发生的是fscanf函数写入给定的位置（在您的情况下，它通过值指向的任何位置） > i，而不是写入i 的位置。这可能会以各种令人讨厌的方式破坏你的记忆 - 在你的情况下，在程序中运行＆＃34;运行＆＃34;在崩溃前相当长的时间。

正如@Brandin指出的那样，你的代码还有一个问题（虽然它不太可能成为你问题的根源）。当您尝试打开文件时，您应该始终检查您是否成功。你可以这样做：

#include <assert.h> // at the top of the program // attempt to open the file: fp = fopen("keimeno.txt","r"); // and check whether you succeeded: assert(fp != NULL); // this says "check fp is not NULL. Otherwise, quit."

或者，您可以通过以下方式使事情变得更漂亮：

const char *fileName = "keimeno.txt"; const char *mode = "r"; if((fp=fopen(fileName, mode))==NULL) { printf("cannot open file %s\n", fileName); return -1; }

放置＆＃34;硬连线值＆＃34;几乎总是一个好主意。在程序开头附近，而不是将它们嵌入到函数调用中。

Answer 2

与我的问题相同。我清空列表，并添加了一个新值，该值导致我用另一个值输入了该值，并以错误“ process-exited-with-return-value-3221225477”结束

#include <iostream>
#include <stdlib.h>

using namespace std;

struct Node {
   int data;
   struct Node *left;
   struct Node *right;
};
int count = 0;
struct Node*front = NULL;
struct Node*rear = NULL;

void enqueue(int val) {
    count++;
   struct Node* newnode = new Node; //= (struct Node*) malloc(sizeof(struct Node));
   newnode->data = val;
   if (front == NULL) {
        front = rear = newnode;
        newnode->right=newnode->left;
        newnode->left=newnode->right;
   }
   else {
    rear->right = newnode;
    newnode->left = rear;
    rear = newnode;
    front->left = rear;
    rear->right = front;
   }
}

void dequeue() {
   if(front==NULL)
      cout<<"There is no element yet. Cannot be dequeue."<<endl;
   else {
      if (front == rear) {
        cout<<"The dequeued element is "<< front->data <<endl;
        front = rear = NULL;
         delete front;
         }
      else {
      cout<<"The dequeued element is "<< front->data <<endl;
       struct Node *curr = front;
       front = front->right;
       front->left = rear;
       rear->right = front;
       delete curr;

       }
   }
}

void display() {
   int i;
   struct Node* ptr;

   if(front==NULL) {
      cout<<"The List is empty, nothing to display.";
      cout<<endl;
      return;
  }    
      ptr = front;
      cout<<"Element/s : ";
      for (i=0;i<count; i++) {
         cout<< ptr->data <<" ";
         ptr = ptr->right;
     }
        if (ptr->right==rear) {
            cout<< ptr->data << " ";
        }
       cout<<endl;
}

int main() {
   int ch, val;
   cout<<"1) Enqueue"<<endl;
   cout<<"2) Dequeue"<<endl;
   cout<<"3) Display"<<endl;
   cout<<"4) Exit"<<endl;
   do {
      cout<<"Enter your choice: "<<endl;
      cin>>ch;
      switch(ch) {
         case 1: {
            cout<<"Enter the value of the element:"<<endl;
            cin>>val;
            enqueue(val);
            cout<<endl;
            return main();
            break;
         }
         case 2: {
            dequeue();
            cout<<endl;
            return main();
            break;
         }
         case 3: {
            display();
            cout<<endl;
            return main();
            break;
         }
         case 4: {
            cout<<"Exit"<<endl;
            break;
         }
         default: {
            cout<<"Invalid Choice"<<endl;
         }
      }
   }while(ch!=4);
      return 0;
}

Answer 3

exited_return_value-3221225477是一个高度抽象的代码错误放错位置编译器由于各自编译器的gnu逻辑错位而产生相同的代码在另一个系统中执行。超过该值称为系统call（）错误...

进程退出，返回值为3221225477

3 个答案: