删除char指针时出现分段错误

Question

I'm posting two fragments here. The first one is giving me Segmentation Fault on deallocating the memory. Second one is working fine.

1)

int main()
{
  char* ch = new char;
  ch = "hello";
  cout << "\n " << ch << endl;
  delete[] ch;  ////OR delete ch; ---> have tried both
  return 0;
}

2)

int main()
{
  char* ch = new char;
  cin >> ch;
  cout << "\n " << ch << endl;
  delete[] ch;   ///OR delete ch /// Both are working fine....
  return 0;
}

Could anybody please tell me why the first one is failing with Segmentation Fault and second one is working fine with both delete and delete[] . Because to me both the program seems to same.

Answer 1

new char generates exactly 1 character (not an array of 1 character, use new char[1] for that)

so delete[] doesn't apply

in the first example, you overwrite your pointer to your allocated 1 character with a pointer to the character string "hello" - deleting this string (as it is static memory) will result in sesgfault

Edit

int main()
{
    char* ch = new char; // ch points to 1 character in dynamic memory
    ch = "hello";        // overwrite ch with pointer to static memory "hello"
    cout<<"\n "<<ch<<endl; // outputs the content of the static memory
    delete[] ch;         // tries to delete static memory
    return 0;
}

Answer 2

There are issues with both examples:

  char* ch = new char;`
  ch = "hello";`    

The new returns an address that points to dynamically allocated memory. You must save this return value so that delete can be issued later. The code above overwrites this value with "hello" (a string-literal). You now have lost the value, and thus can not call delete with the proper value.

The second example, even though you say "works fine" is still faulty.

  char* ch = new char;`    
  delete[] ch;   ///OR delete ch /// Both are working fine....`    

The wrong form of delete is used. You allocated with new , so you must deallocate with delete , not delete[] . It works this way: new -> delete , new[] -> delete[] .

Unlike most other languages, if you go against the rules of C++, corrupt memory, overwrite a buffer, etc., there is no guarantee that your program will crash, seg fault, etc. to let you know that you've done something wrong.

In this case, you're lucky that simple types such as char* are not affected by you using the wrong form of delete . But you cannot guarantee that this will always work if you change compilers, runtime settings, etc.

Answer 3

In the first one, you change the pointer to point to a string literal:

ch = "hello";

String literals are static arrays, so mustn't be deleted.

The second is wrong for at least two reasons:

• you allocate a single character, not an array; a single character would be deleted with delete not delete[]
• cin>>ch will (most likely) read more than one character, but you've only allocated space for one.

Both of these cause undefined behaviour, which might manifest itself as a visible error, or might appear to "work fine" - but could fail when you least expect it.

To allocate an array, use new char[SIZE] ; but even then, you can't prevent the user from giving too much input and overflowing the buffer.

Unless you're teaching yourself how to juggle raw memory (which is a dark art, best avoided unless absolutely necessary), you should stick to high-level types that manage memory for you:

std::string string;
string = "hello";
std::cout << string << '\n';
std::cin >> string;
std::cout << string << '\n';

Answer 4

There are a couple of problems with each, namely that you're only allocating a single character when you're trying to allocate a character array.

In the first example, you're also allocating a single character and then subsequently reassign the pointer to a character array - ch = "hello" will not copy the string, just reassign the pointer. Your call to delete[] will then attempt to delete a string that is not heap allocated, hence the seg fault. And you're also leaking the char you allocated, too.

Answer 5

Here's what's wrong with both snippets:

First snippet:

char* ch = new char; ch = "hello";

It's not legal to assign a string literal to a non-const char pointer . Also, you re-assign the pointer immediately after you call new . The original value returned by new is now lost forever and can not be free for the duration of the program. This is known as a memory leak.

delete[] ch;

You try to deallocate the string literal. This crashes your program. You are only allowed to delete pointers that you get from new and delete[] pointers that you get from new[] . Deleting anything else has undefined behaviour.

Second snippet:

cout<<"\n "<<ch<<endl;

ch points to a single character, not a zero terminated char array . Passing this pointer to cout has undefined behaviour. You should use cout << *ch; to print that single character or make sure that ch points to a character array that is zero terminated.

delete[] ch;

You allocated with new , you must deallocate with delete . Using delete[] here has undefined behaviour.

Both are working fine....

"working fine" is one possible outcome of undefined behaviour, just like a runtime error is.

To answer the question, neither snippet is correct. First one crashes because you got lucky, second one appears to work because you got unlucky.

Solution: Use std::string .

Answer 6

there are several errors in your programs.

In the first program you are not deleting something dynamically allocated but the statically allocated string "hello". Infact when you execute ch="hello" you are not copying the string in the wrongly allocated buffer "new char" ( this new just allocates one char, not what you are looking for ) but you makes the pointer ch to point to the start of the string "hello" located somewhere in the non writable memory ( normaly that string are pointed directly into the executable ). So the delete operation is trying to deallocate something that cannot be deallocate. So the first program culd be rewritten like:

int main()
{
  const char* ch = "hello";
  cout<<"\n "<<ch<<endl;
  return 0;
}

or like

int main()
{
  char* ch = new char[strlen("hello")+1];
  strcpy( ch, "hello");
  cout<<"\n "<<ch<<endl;
  delete[] ch;  // with square brackets, it's an array
  return 0;
}

Answer 7

You should use something like:

char* ch = new char[6] ;

strcpy(ch,"hello") ;

...

delete[] ch ;