Dealing with chars in C++ (without std::string)

Question

I have this code:

char* value = "abcdefg";

char* secondValue = value;

The second value will get the addres of value ok? If I delete "value" the secondValue won't be available am I right?

so I should do:

char* value = "abcdefg";
secondValue = new char[strlen(value)];
strcpy(secondValue, value);

so If I delete "value" no problem.

And finally to dealloc the secondValue I should do:

delete[] secondValue;

am I right?

Your questions is tagged and titled C, but `delete` is C++ syntax. Your question is a bit muddled. — Nathan Fellman, Nov 20 '10 at 19:25
new/delete? That's C++. Perhaps you should look at std::string. — Etienne de Martel, Nov 20 '10 at 19:26
ok I'll change the title, I do not have access to std stuff on the component I'm using. — okami, Nov 20 '10 at 19:27
You should replace `char*` with `const char*`. The reason is outlined [here](http://stackoverflow.com/questions/4224682/standard-conversions-array-to-pointer-conversion-strings/4224732#4224732). — fredoverflow, Nov 20 '10 at 20:09

score 6 · Answer 1 · answered Nov 20 '10 at 19:25

There are two issues with what you wrote:

You cannot delete char *value = "abcdefg";, as it is not allocated on the heap. To allocate heap memory you use new (in C++) or malloc (in C).
When you allocate memory for a string, you always need one more extra byte for the null termination.

In your case, you should have done:

secondValue = new char[strlen(value)+1];

Other than that, you are correct

score 3 · Answer 2 · answered Nov 20 '10 at 19:27

If you are using C++ you are correct, except that you need to make secondValue one character bigger:

secondValue = new char[strlen(value) + 1];

C-style strings are terminated with a '\0' character, which also need space to be stored.

If you are using C instead of C++, there is no new or delete[] and you have to use the functions malloc() and free() instead:

secondValue = malloc(strlen(value) + 1);
...
free(secondValue);

In any case note that in the example value is a string literal, which cannot be deleted/freed. You should only delete/free what you allocated with new/malloc respectively.

score 1 · Answer 3 · answered Nov 20 '10 at 19:28

1

The way you wrote your code:

char* value = "abcdefg";

The compiler will generate a static string "abcdefg", and value will be a pointer to that. You may assign secondvalue = value and then have value go out of scope, andsecondvalue` will still be valid.

In your example there is no other way to deallocate value.

answered Nov 20 '10 at 19:28

Nathan Fellman

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so if the char* value = "abcdefg"; is inner a struct like struct { char* value; } MyStruct; the "value" data will be a static one even if I delete a object of MyStruct type? – okami Nov 20 '10 at 19:32
Yes, **if** you initialized `value` with a static constant. – Nathan Fellman Nov 20 '10 at 19:37

Dealing with chars in C++ (without std::string)

3 Answers3