c++ - 在 C++ 中使用指向 char 的指针时没有返回错误 1 ​​的输出

Question

我已经彻底寻找答案无济于事。我一直在使用 Starting out with C++ Gaddis 教科书作为引用。我知道这个问题有一个简单的答案，但我很难弄明白。这是我的代码:

class Employee
{
    private:
    char *name;
    int idNumber;
    char *department;
    char *position;
    void initName(const char *n)
{
    name = new char[strlen(n) + 1];
    strcpy(name, n);
}
void initDepartment(char *d)
{
    department = new char[strlen(d) + 1];
    strcpy(department, d);
}

void initPosition(char *p)
{
    position = new char[strlen(p) + 1];
    strcpy(position, p);
}

public:

Employee()
{
    strcpy(name, "");
    idNumber = 0;
    strcpy(department, "");
    strcpy(position, "");

}
Employee(char *nam, int num, char *dep, char *posit)
{
    initName(nam);
    idNumber = num;
    initDepartment(dep);
    initPosition(posit);
}

Employee(const char *nam, int num)
{
    initName(nam);
    idNumber = num;
    strcpy(department, "");
    strcpy(position, "");   
}

~Employee()
{
    delete [] name;
    delete [] department;
    delete [] position;
}

void setName(char *n)
{
    name = new char[strlen(n) + 1];
    strcpy(name, n);
}

void setIdNumber(int num)
{
    idNumber = num;
}

void setDepartment(char *d)
{
    department = new char[strlen(d) + 1];
    strcpy(department, d);
}

void setPosition(char *p)
{
    position = new char[strlen(p) + 1];
    strcpy(position, p);
}

const char * getName() const
{
    return name;
}

int getIdNumber() const
{
    return idNumber;
}

const char * getDepartment() const 
{
    return department;
}

const char * getPosition() const 
{
    return position;
}
};

int main(int argc, char** argv) 
{
const int SIZE = 50;
const char name[SIZE] = "Mark Jones";
Employee employee(name, 3452);
const char *ptr = NULL;

ptr = employee.getName();

cout << ptr << endl;

return 0;
}

Answer 1

在默认构造函数中，您将空字符串复制到垃圾指针:

Employee()
// name and all other members are not initialized
{
    strcpy(name, ""); // << copy to unallocated memory
    idNumber = 0;
    strcpy(department, ""); // << and here
    strcpy(position, ""); // << and here
}

正确的默认构造函数是:

Employee()
: name(nullptr) // Initialize all members here
, idNumber(0)
, department(nullptr)
, position(nullptr)
{
    // Initialize c-strings
    initName("");
    initDepartment("");
    initPosition("");
}

不要忘记一直使用const。以及类似的其他构造函数:

Employee(const char* nam, int num, const char* dep, const char* posit)
: name(nullptr) // << Initialize all members here
, idNumber(num) // << Initialize number with proper value
, department(nullptr)
, position(nullptr)
{
    initName(nam);
    initDepartment(dep);
    initPosition(posit);
}

Employee(const char* nam, int num)
: name(nullptr) // Initialize all members here
, idNumber(num) // << Initialize number with proper value
, department(nullptr)
, position(nullptr)
{
    initName(nam);
    initDepartment("");
    initPosition("");   
}

适当的函数参数:

void initName(const char *n)
{
    name = new char[strlen(n) + 1];
    strcpy(name, n);
}
void initDepartment(const char *d)
{
    department = new char[strlen(d) + 1];
    strcpy(department, d);
}

void initPosition(const char *p)
{
    position = new char[strlen(p) + 1];
    strcpy(position, p);
}

并且不要忘记清除 setter 中的内存(否则就是内存泄漏):

void setName(const char *n)
{
    delete[] name;
    name = new char[strlen(n) + 1];
    strcpy(name, n);
}

void setDepartment(const char *d)
{
    delete[] department;
    department = new char[strlen(d) + 1];
    strcpy(department, d);
}

void setPosition(char *p)
{
    delete[] position;
    position = new char[strlen(p) + 1];
    strcpy(position, p);
}

这应该可以修复您所有的运行时错误。