Returning char*

If i create a function:

Code:

---

char *fn1(){
  char *retVal=new char[100];
  strcpy(retVal, "test");
  return(retVal);
}

---

And use it like this:

Code:

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char *ptr=0;
if(ptr=fn1())
  printf(ptr);
delete [] ptr;

---

Is this the correct way to return a pointer to char?
Does the compiler know the amount of allocated memory, or am i messing everything up?

Hi,

In the current standard, if new failed to allocate memory, it is supposed to raise an exception and not the old way of returning 0.

If we assume that your compiler returns 0 if new fails, in that case, it seems correct.

You did it correctly.

It's right, unless the dynamic memory allocation fails... Which also doesn't mean that the code isn't correct.

Thx guys.
I was just insecure regarding the deletion of allocated memory...

You code is OK, but I would not return a string like this.

The code calling the fn1 will have a requirement to delete[] the string array. This is often considered bad practice. There are better ways to return strings in C++.

You could:

1. Return a string object. This is best. I.e.

Code:

---

string fn1()
{
  return "test";
}

---

2. If you want you're code to use char*, you could cause fn1 to copy the result into an existing buffer.

Code:

---

void fn1(char* s, int s_len)
{
  strncpy(s, "test", s_len);
}

---

3. There are other ways also, including the one you suggested. But whether you should use them depends on the requirements of your project. I.e. if your code forms a DLL etc.

Code:

---

void fn1(char* s, int s_len)
{ 
    strncpy(s, "test", s_len);
}

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Don't forget, this isn't guaranteed to NULL terminate the array. Only if there is room in the array will strncpy include the \0.

Code:

---

#include <new>

// ...
char *retVal = new(nothrow) char[100];
// will be NULL on failed alloc

---

Use new(nothrow) if you want 0 returned on failure (instead of throwing std::bad_alloc)