char [length] initialization and dealing with it

Question

I have defined a char array:

char d[6];

Correct me if I'm wrong regarding following:

At this moment no memory is allocated for variable d. Now I'm going to initialize it:

d="aaaaa";

After this kind of initialization, there would be no need to free memory; it will be done automatically.

How do I know if the char[] was initialized? I am looking for a pattern like

if (filled(d)){..}

Also, how do I fill char[] with one kind of character?

Answer 1

At this moment no memory allocated for variable d.

Incorrect. This:

char d[6];

is an uninitialised array of 6 chars and memory, on stack, has been allocated for it. Stack variables do not need to be explicitly free()d, whether they are initialised or not. The memory used by a stack variable will be released when it goes out of scope. Only pointers obtained via malloc(), realloc() or calloc() should be passed to free().

To initialise:

char d[6] = "aaaaa"; /* 5 'a's and one null terminator. */

or:

char d[] = "aaaaa"; /* The size of the array is inferred. */

And, as already noted by mathematician1975, array assignment is illegal:

char d[] = "aaaaa"; /* OK, initialisation. */
d = "aaaaa";        /* !OK, assignment. */

strcpy(), strncpy(), memcpy(), snprintf(), etc can be used to copy into d after declaration, or assignment of char to individual elements of d.

---

How to know was char[] initialized? I need pattern if filled(d){..}

If the arrays are null terminated you can use strcmp()

if (0 == strcmp("aaaaaa", d))
{
    /* Filled with 'a's. */
}

or use memcmp() if not null terminated:

if (0 == memcmp("aaaaaa", d, 6))
{
    /* Filled with 'a's. */
}

---

How to fill char[] with one kind of characters?

Use memset():

memset(d, 'a', sizeof(d)); /* WARNING: no null terminator. */

or:

char d[] = { 'a', 'a', 'a', 'a', 'a', 'a' }; /* Again, no null. */

Answer 2

Your code will not compile (gcc 4.6.3) if you do

 char d[6];
 d = "aaaaa";

you will need to do

 char d[6] = "aaaaa" 

to initialise it this way. This is a local variable created on the stack and so in terms of memory issues all you need worry about is not writing/reading beyond the array bounds.

Answer 3

First, whenever you declare char d[6] 6 bytes of memory is already allocated.

Second, no need to free your memory unless you do malloc

Third, if you want to initialize it with one kind of character then do this

char d[6] = "aaaaa"; 

Answer 4

int d[6];

6 bytes will be allocated onto the stack with this declaration. It will be freed automatically.

Answer 5

How to know was char[] initialized? I need pattern if filled(d){..}

Just do this while declaring the character array:

char d[6];
d[0] = 0;

Then you can check like this:

if(strlen(d) == 0)
//d is not initialized