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 variabl

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.

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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. */
}

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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. */