C++ 进程在 4.017 秒后退出，返回值 3221225477

Question

我是 C++ 新手，我正在尝试编写这个简单的代码来输入和显示员工数据。 我想通过地址将用户输入传递给 inputEmployee 函数，将其保存在数组中然后显示。当我进入输入员工姓名的部分时，我收到以下消息“进程在 4.017 秒后退出，返回值 3221225477”。

有人可以提出解决方案或解释我的代码有什么问题吗？

我不确定我所做的是否正确。

struct Employee {
    char    name[30];
    char    surname[30];
    int     age;
    float   salary;
};

Employee* inputEmployee(struct Employee *emp, int n, int index) {
     Employee *employeeRecords = new Employee[n];
     employeeRecords[index] = *emp;
     return employeeRecords;
}

int displayEmployee(struct Employee *arr) {
    int arraySize = sizeof(arr)/sizeof(*arr);
    printf("\nEmployee data:");
    for(int i=0; i < arraySize; i++) {
        printf("Name: %s", arr[i].name);
        printf("Surname: %d", arr[i].surname);
        printf("Age: %d", arr[i].age);
        printf("Salary: %f\n", arr[i].salary);
    }
    return 0;
}

int main()
{
    int n = 0;
    printf("\nEnter employee number :\n");
    scanf("%d", &n);
    struct Employee *eArr;

    for(int i=0; i<n; i++) {
        struct Employee *emp = NULL;

        printf("\nEnter employee data :\n");
        printf("Name ?:"); scanf("%s", emp->name);
        printf("Surname ?:"); scanf("%s", emp->surname);
        printf("Age ?:"); scanf("%d", &emp->age);
        printf("Salary ?:"); scanf("%f", &emp->salary);
        eArr = inputEmployee(emp, n, i);
    }

    displayEmployee(eArr);

    return 0;
}

Answer 1

您应该退后一步，阅读有关 C++ 中的内存管理、数组、指针和引用的内容。当你想编写 C++ 时，这是一个很难的部分，你必须学习。您绝对应该阅读更多文章。

边做边学只是成功的一半。您需要了解在后台发生了什么。您的程序可以做“某事”，即使它看起来不错，但以后可能会严重失败。但请保持警惕，继续前进。

---

Employee *emp = NULL; 在你的内存中创建一个指向“无处”的指针。您至少要做的事情是Employee *emp = new Employee();。这将分配可用于数据的内存。请记住，您必须为每个分配的内存使用delete 运算符。

我试图修复您的代码。我为整个数组分配了内存。

#include <cstdio>

struct Employee
{
  char name[30];
  char surname[30];
  int age;
  float salary;
};

//--------------------------------------------------------------------------//
int displayEmployee(Employee *arr, int count)
{
  printf("\nEmployee data:");
  for (int i = 0; i < count; i++)
  {
    printf("Name: %s ", arr[i].name);
    printf("Surname: %s ", arr[i].surname);
    printf("Age: %d ", arr[i].age);
    printf("Salary: %f\n", arr[i].salary);
  }
  return 0;
}

//--------------------------------------------------------------------------//
int main()
{
  int n = 0;
  printf("\nEnter employee number: \n");
  scanf("%d", &n);

  Employee *eArr = new Employee[n];
  for (int i = 0; i < n; i++)
  {
    Employee &emp = eArr[i];

    printf("\nEnter employee data :\n");
    printf("Name ?: ");
    scanf("%s", emp.name);
    printf("Surname ?: ");
    scanf("%s", emp.surname);
    printf("Age ?: ");
    scanf("%d", &(emp.age));
    printf("Salary ?: ");
    scanf("%f", &(emp.salary));
    // eArr = inputEmployee(emp, n, i);
  }
  displayEmployee(eArr, n);

  delete[] eArr;
  return 0;
}

Answer 2

您已经有了一个很好的答案，但您还必须涵盖几个额外的点，以便对您的输入程序有信心。首先，您必须验证每个用户输入，并且必须在下一次输入之前清空 stdin 的任何其他字符。问问你自己，如果你输入"Ninety One" 换成Age，会发生什么？清除stdin 可以通过一个简单的循环继续您的C-stye 输入或使用std::cin.ignore()（如果使用iostream）来完成。

例如，当您循环收集员工数据时，您必须强制用户为每个变量输入有效的输入，然后才能进入下一个商店，例如

    for (int i = 0; i < n; i++) {   /* loop for input of n empoloyees */
        struct Employee emp;

        puts ("\nEnter employee data :");
        for (;;) {  /* loop continually until valid input received */
            fputs ("\n  Name ?: ", stdout);     /* prompt (fputs for \n control) */
            int rtn = scanf("%s", emp.name);    /* save return from scanf */
            if (rtn == EOF) {                   /* check for manual EOF */
                puts ("(user canceled input)");
                return 0;
            }   /* remove any extraneous characters from stdin */
            for (char c = getchar(); c != '\n' && c != EOF; c = getchar()) {}
            if (rtn == 1)                       /* if good input break loop */
                break;
        }

您必须在处理数字输入时添加额外的检查以涵盖 匹配失败 情况，例如

        for (;;) {  /* ditto for Age, but note the additional validation required */
            fputs ("  Age ?: ", stdout);
            int rtn = scanf ("%d", &emp.age);
            if (rtn == EOF) {
                puts ("(user canceled input)");
                return 0;
            }
            for (char c = getchar(); c != '\n' && c != EOF; c = getchar()) {}
            if (rtn == 1)
                break;
            fputs ("error: invalid integer input.\n", stderr);
        }

如果您要求输入的数字在特定范围内，您可以添加额外的检查，例如 if (emp.age < 16 || emp.age > 80) 处理无效年龄的错误。

对于输出，您不需要多次调用printf（或std::cout）——一个就可以了，例如

/* No return, you must pass number of elements as second parameter */
void displayEmployee (struct Employee *arr, int n)
{
    puts ("\nEmployee data:");      /* no conversion, just use puts */

    for (int i=0; i < n; i++)       /* only 1 printf needed */
        printf ("\n  Name: %s\n  Surname: %s\n  Age: %d\n  Salary: %.2f\n",
                arr[i].name, arr[i].surname, arr[i].age, arr[i].salary);
}

还请注意，没有理由从displayEmployee 返回值，它只是一个输出函数，没有任何返回来指示函数内操作的成功或失败。

总而言之，你可以这样做：

#include <cstdio>

#define MAXN 30

struct Employee {
    char    name[MAXN];
    char    surname[MAXN];
    int     age;
    float   salary;
};

/* No return, you must pass number of elements as second parameter */
void displayEmployee (struct Employee *arr, int n)
{
    puts ("\nEmployee data:");      /* no conversion, just use puts */

    for (int i=0; i < n; i++)       /* only 1 printf needed */
        printf ("\n  Name: %s\n  Surname: %s\n  Age: %d\n  Salary: %.2f\n",
                arr[i].name, arr[i].surname, arr[i].age, arr[i].salary);
}

int main()
{
    int n = 0;
    fputs ("\nEnter employee number : ", stdout);
    if (scanf ("%d", &n) != 1) {                    /* validate every input */
        fputs ("error: invalid integer input.\n", stderr);
        return 1;
    }
    struct Employee *eArr = new struct Employee[n]; /* allocate */

    for (int i = 0; i < n; i++) {   /* loop for input of n empoloyees */
        struct Employee emp;

        puts ("\nEnter employee data :");
        for (;;) {  /* loop continually until valid input received */
            fputs ("\n  Name ?: ", stdout);     /* prompt (fputs for \n control) */
            int rtn = scanf("%s", emp.name);    /* save return from scanf */
            if (rtn == EOF) {                   /* check for manual EOF */
                puts ("(user canceled input)");
                return 0;
            }   /* remove any extraneous characters from stdin */
            for (char c = getchar(); c != '\n' && c != EOF; c = getchar()) {}
            if (rtn == 1)                       /* if good input break loop */
                break;
        }
        for (;;) {  /* ditto for Surname */
            fputs ("  Surname ?: ", stdout);
            int rtn = scanf("%s", emp.surname);
            if (rtn == EOF) {
                puts ("(user canceled input)");
                return 0;
            }
            for (char c = getchar(); c != '\n' && c != EOF; c = getchar()) {}
            if (rtn == 1)
                break;
        }
        for (;;) {  /* ditto for Age, but note the additional validation required */
            fputs ("  Age ?: ", stdout);
            int rtn = scanf ("%d", &emp.age);
            if (rtn == EOF) {
                puts ("(user canceled input)");
                return 0;
            }
            for (char c = getchar(); c != '\n' && c != EOF; c = getchar()) {}
            if (rtn == 1)
                break;
            fputs ("error: invalid integer input.\n", stderr);
        }
        for (;;) {  /* ditto for Salary with same note */
            fputs ("  Salary ?: ", stdout);
            int rtn = scanf ("%f", &emp.salary);
            if (rtn == EOF) {
                puts ("(user canceled input)");
                return 0;
            }
            for (char c = getchar(); c != '\n' && c != EOF; c = getchar()) {}
            if (rtn == 1)
                break;
            fputs ("error: invalid floating-point input.\n", stderr);
        }
        eArr[i] = emp;  /* simply assign struct to element */
    }

    displayEmployee(eArr, n);

    return 0;
}

使用/输出示例

$ ./bin/empstruct

Enter employee number : 3

Enter employee data :

  Name ?: Mickey
  Surname ?: Mouse
  Age ?: 91
  Salary ?: 88.82

Enter employee data :

  Name ?: Minnie
  Surname ?: Mouse
  Age ?: 88
  Salary ?: 93.25

Enter employee data :

  Name ?: Goofy
  Surname ?: Dog
  Age ?: 85
  Salary ?: Bone and Treats
error: invalid floating-point input.
  Salary ?: .25

Employee data:

  Name: Mickey
  Surname: Mouse
  Age: 91
  Salary: 88.82

  Name: Minnie
  Surname: Mouse
  Age: 88
  Salary: 93.25

  Name: Goofy
  Surname: Dog
  Age: 85
  Salary: 0.25

尝试使用您当前的代码为Age 或Salary 输入无效的输入，然后看看会发生什么（当您这样做时，请将您的手指悬停在 Ctrl+c 上...或查看哪个输入似乎被跳过）

此外，您应该选择 C ​​或 C++。您所拥有的是一个使用new 而不是malloc 进行分配的C 程序。如果你用new 替换你的分配struct Employee *eArr = malloc (n * sizeof *eArr); 那么你有一个纯C 程序。语言混用没有什么不好的，但就是不好看。

如果您还有其他问题，请告诉我。