Thread: difference between char *var and char var[]

Originally Posted by liri

would this break?

Yes, in fact, it wouldn't even compile...

Code:

char var1[80];  //I have enough memory for 80 chars
char *var2;      //I hold the address to a char somewhere

You cannot assign a string in quotes to a char array, you'll need to use snprintf.
Note that var2 has absolutely *no* memory allocated whatsoever (except for itself). So trying to store a string in var2 will crash your program. Var2 *points* to a bit of memory. To use var2, it either has to point to some other char array (like var1) or to memory that you have allocated on the heap.

QuantumPete

Originally Posted by liri

Hey,

If I'm not mistaken it is possible to specify both ways and they are identical
although is this valid for assignments of variables later?

for example:

char var[] = "test";
var = "testing more options";

would this break?
when var[] initializes it reserves enough memory, as much as "test" but later a longer string is evaluated to var, isn't this bad?


thanks.

The compiler should not allow this.

The difference is that char *var = "test" will get the address of a constant string stored somewhere in your executable file, char var[] = "test" will make some memory available that is filled in with "test". The latter can't be assigned a different value (but you can modify the content, and if you want to use strcpy() to fill it in, then you can do that). On the other hand, the pointer form can easily be changed to point to a different string, but if it's a string constant, it shouldn't be modified.

If you want to do the above and be able to modify the string too, then I would suggest this:

Code:

char var[100] = "test";
...
  strcpy(var, "testing more options");
...

If all you want is to have different strings depending on some condition, and don't plan to modify the string, you can use char *var, it would be the better approach.

--
Mats

If you do char var[] = "", then you get an array of a single byte, which is quite useless.

If you really want to do that, you should give a length.

--
Mats

Yes, in theory it is. But normally if you use the "" initialiser with empty brackets, you're basically saying to anyone reading your code: "This variable is going to hold "test" and nothing else". If you then re-assign it, it's going to confuse a lot of people.

QuantumPete

Originally Posted by liri

Well, I'm talking about converting this form to character array:
char *var = "test";
...code...
var = "";

so doing that with strcpy(var, "") seems like the normal way of doing it.
unless you can propose something else


regards.

That is what you'd have to do - but as QuantumPete says - it is not really how one would recommend doing it.

The way I would recommend doing it is something like this:

Code:

   char var[10];  // Suitable length should be choosen based on what you need. 
   strcpy(var, "");
...code...
   if (somecondition)
      strcpy(var, "Text");
....
   if (othercondition)
      strcpy(var, ...);

[Using strncpy or snprintf may be better choices, as those are "safe" to prevent the string from being too long.]

--
Mats