Question

我有一份家庭作业，但我不明白我的代码有什么问题。名称变量搞砸了。我查看了我的教科书，但我的代码看不出任何问题。请帮忙。

名称必须以80的数组长度存储，这是我知道如何做到这一点的唯一方式。

我来自很多java经验，所以我不知道是否存在我缺少的语法或什么。

感谢您的时间。

//import statements
#include <iostream>
#include <iomanip>
using namespace std;

//class declaration section
class Student
{
    //declare private instance variables
    private:
        int ssn;
        char* name[80];
        const int ARRAYLENGTH =80;

    //deckale public methods and constructor
    public:
        Student();
        void setName(string);
        int getSSN();
        string getName();
        void setSSN(int);


};

//class implementation section
Student::Student()
{
    //set ssn equal to 99999999 by default
    ssn = 999999999;
    //set name equal to unassigned by default
    string someString ="unassigned";
    for(int i = 0; i < ARRAYLENGTH; i++)
    {
        if (i < someString.length())
            name[i] = &someString[i];
        else
            name[i]= "";
    }
}

void Student::setName(string newName)
{
    //assigns the newName to the char array
    for(int i = 0; i < ARRAYLENGTH; i++)
    {
        if (i < newName.length())
            name[i] = &newName[i];
        else
            name[i]= "";
    }

    return;
}

int Student::getSSN()
{
    return ssn;
}

string Student::getName()
{
    //make a string from the char array to return a string
    string name1 = "";
    for(int i = 0; i < 80; i++)
    {
        name1 += *(name+i);
    }
    return name1;
}

void Student::setSSN(int num)
{
    //check to make sure ssn isnt below or equal to 0
    if (num > 0)
        ssn = num;

    return;
}


//main function
int main() {

    Student student1, student2;

    //changing name and ssn of student2
    student2.setName("John Doe");
    student2.setSSN(123456789);

    //printing out information
    cout<< "Name for student1 is "<< student1.getName() <<" and snn is "<< student1.getSSN() <<endl; //name should be unassigned and ssn should be 999999999
    cout<< "Name for student1 is "<< student2.getName() <<" and snn is "<< student2.getSSN() <<endl; //name should be John Doe and ssn should be 123456789

    return 0;

}

Answer 1

这个char* name[80]声明了一个指针数组。也许您只需要使用char name[80]来获取char数组。（这实际上是指向数组的第一个字符的指针）

char* name[80]是一个指针数组。（可能不是你想要的）

Answer 2

我在单次扫描中指出的一些错误是： -

这是错误的： -

name[i] = &someString[i];

您正在尝试在字符串中获取特定字符的地址，然后将其放在char *。

然后，您尝试将值存储在int范围之外： -

ssn = 999999999;

Answer 3

您正在更改Student2的SSN，但您正在打印student1的SSN

Answer 4

此代码应该可以解决问题。

//import statements
#include <iostream>
#include <iomanip>
using namespace std;

//class declaration section
class Student
{
    //declare private instance variables
    private:
        int ssn;
        char name[80];
        const int ARRAYLENGTH =80;

    //deckale public methods and constructor
    public:
        Student();
        void setName(string);
        int getSSN();
        string getName();
        void setSSN(int);


};

//class implementation section
Student::Student()
{
    //set ssn equal to 99999999 by default
    ssn = 999999999;
    //set name equal to unassigned by default
    string someString ="unassigned";
    for(int i = 0; i < ARRAYLENGTH; i++)
    {
        if (i < someString.length())
            name[i] = someString[i];
        else
            name[i]= '\0';
    } 
}

void Student::setName(string newName)
{
    //assigns the newName to the char array
    for(int i = 0; i < ARRAYLENGTH; i++)
    {
        if (i < newName.length())
            name[i] = newName[i];
        else
            name[i]= '\0';
    }

    return;
}

int Student::getSSN()
{
    return ssn;
}

string Student::getName()
{
    //make a string from the char array to return a string
    string name1 = "";
    for(int i = 0; i < 80; i++)
    {
        name1 += *(name+i);
    }
    return name1;
}

void Student::setSSN(int num)
{
    //check to make sure ssn isnt below or equal to 0
    if (num > 0)
        ssn = num;

    return;
}


//main function
int main() {

    Student student1, student2;

    //changing name and ssn of student2
    student2.setName("John Doe");
    student2.setSSN(123456789);

    //printing out information
    cout<< "Name for student1 is "<< student1.getName() <<" and snn is "<< student1.getSSN() <<endl; //name should be unassigned and ssn should be 999999999
    cout<< "Name for student1 is "<< student2.getName() <<" and snn is "<< student2.getSSN() <<endl; //name should be John Doe and ssn should be 123456789

    return 0;

}

Answer 5

我真的不认为你要存储80个名字，看起来学生只有一个名字。

char *name[80]表示80个名字，而不是80个字符！

无论如何：你不能保持指向局部变量的指针，你在许多地方都这样做。

因此，您最好使用const char*而不是值string。例如：

将string someString ="unassigned";更改为const char* someString = "unassigned"

函数Student::setName(string newName)应为Student::setName(const char* newName)

另一个选项是复制，而不仅仅是指定指针。表示使用string name[80];代替char* name[80]; 在这种情况下：函数Student::setName(string newName)最好是Student::setName(const string& newName)

c ++内部类，方法不起作用

5 个答案: