Passing character pointers into a function and dynamically allocating memory

Question

Student.h

class Student
{
 private:
      char m_sHouse[64];
 public:
 Student(void);
 ~Student(void);
 void getHouse(char *hName);
 void setHouse(char *hName);
}

Student.cpp

 void Student::setHouse(char *hName)
 {
    strcpy(m_sHouse, hName);
 }

 void Student::getHouse(char *hName)
 {
     if (m_sHouse != NULL)
     {
        hName = new char[strlen(m_sHouse)+1];
        strcpy(hName, m_sHouse);
     }
 }

In main:

 student.getHouse(house);
 if (strcmp(house, "house") == 0)
     cout <<"\tCorrectly returned the student house: " << house<< endl;

setHouse(char *hName) sets student->m_sHouse equal to "house".

My question:

When inside getHouse(char *hName), it acts as it should, setting hName to "house". but when control is passed out of the function, my dynamically allocated memory is deallocated, so when I strcmp in main, my program crashes (I end up comparing a NULL pointer).

The given answers will solve this problem, but may I ask why you need to do this? Using `strcpy` in C++ is usually frowned upon, since you're mixing different styles. Have you tried working with `std:string`s? — Alexander Kondratskiy, Jul 02 '12 at 18:50
While it is important for you to understand why your code is crashing, it is better to switch to C++ strings to avoid the problem entirely. — jlunavtgrad, Jul 02 '12 at 18:52
I need to do it this way because my teacher is old. He understands that using strings is the way to go, but the way he sees it, we may have to work with either old code, or old programmers stuck in their ways, so he wants us to be comfortable using kernigan and ritchie character arrays. — Nick, Jul 02 '12 at 18:56
@Nick If you need help, I'd be glad to help out. By the way, this is Rich. — Drise, Jul 02 '12 at 19:16
@Nick Indeed. I can help before/after class, if you have a preferred time. — Drise, Jul 02 '12 at 19:21
either works for me. i'll probably be there early, if thats cool w/ you. thanks man :D — Nick, Jul 02 '12 at 19:23
Also, if someone has answered your question, plea see mark it as such by clicking the check mark next to the answer. — Drise, Jul 02 '12 at 19:50
Unfortunately, we did not set a place to meet. On that note, I'm in the lounge like area right next to the classroom. — Drise, Jul 02 '12 at 21:37

Drise · Accepted Answer · 2012-07-02T19:40:36.927

4

Nick, the proper solution is that you know that hName is already allocated by the user of the class (Dr. Coleman). You simply need to strcpy into the character array.

Simply put:

void Student::getHouse(char *hName)
{
  strcpy(hName, m_sHouse);
}

edited Jul 02 '12 at 19:40

answered Jul 02 '12 at 19:23

Drise

4,310
5
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66

of course its the easy answer....appreciate it mate. other answers were correct, just not following the specific parameters the dr. coleman gave....then again, I didn't specify. see you in a few hours. – Nick Jul 02 '12 at 20:02

score 3 · Answer 2 · answered Jul 02 '12 at 18:43

3

You are allocating new memory and assigning it to a local variable. Change your function to

 void Student::getHouse(char **hName)
 {
     if (m_sHouse != NULL)
     {
        *hName = new char[strlen(m_sHouse)+1];
        strcpy(*hName, m_sHouse);
     }
 }

This will change the address pointed to by the argument passed in to your function, not the copy of it

answered Jul 02 '12 at 18:43

cppguy

3,611
2
21
36

I understand that this will work, however, my teacher gave us extremely specific requirements when naming/defining class functions, because he grades by using his own main. he told us to specifically use the function prototype (if thats even what you call them when its a class function) `void getName(char *mName, char *wName);` so, i cant make a pointer to a pointer. – Nick Jul 02 '12 at 19:10

score 2 · Answer 3 · answered Jul 02 '12 at 18:45

The pointer hname is a copy of house (the pointer you have passed to getHouse). Inside that function you change hname, however, you are not changing the original house! To do that, you should either return the allocated memory:

char *Student::getHouse()
{
    char *hame = NULL;
    if (m_sHouse != NULL)
    {
        hName = new char[strlen(m_sHouse)+1];
        strcpy(hName, m_sHouse);
    }
    return hname;
}

and then

house = student.getHouse();

or give a pointer to this variable, so that it can be changed:

void Student::getHouse(char **hName)
{
    if (m_sHouse != NULL && hname != NULL)
    {
        *hName = new char[strlen(m_sHouse)+1];
        strcpy(*hName, m_sHouse);
    }
}

and then

student.getHouse(&house);

Similarly, you can give a reference to the house variable:

void Student::getHouse(char *&hName)
{
    if (m_sHouse != NULL)
    {
        hName = new char[strlen(m_sHouse)+1];
        strcpy(hName, m_sHouse);
    }
}

and then

student.getHouse(house);

The better solution however, would be to use std::string instead.

score 1 · Answer 4 · answered Jul 02 '12 at 18:43

void Student::getHouse(char *hName)
{
     if (m_sHouse != NULL)
     {
        hName = new char[strlen(m_sHouse)+1];
        strcpy(hName, m_sHouse);
     }
 }

This only modifies a copy of hName pointer and does not modify the original pointer. To modify it you need to pass a pointer to your pointer:

void Student::getHouse(char **hName)
{
     if (m_sHouse != NULL)
     {
        *hName = new char[strlen(m_sHouse)+1];
        strcpy(*hName, m_sHouse);
     }
 }

and call your function like this:

student.getHouse(&house);

Passing character pointers into a function and dynamically allocating memory

4 Answers4