How to access an element of an array inside of an array in c?

Question

I am trying to access the first element inside the array 'levels'.

unsigned char Level1[70] =
{
    0x02,0x02,0x02,0x02,0x02,0x02,0x02,0x02,0x02,0x02,0x02,0x00,0x01,0x01,0x01,0x01,0x01,0x03,0x07,0x02,0x02,0x01,0x01,0x01,0x01,0x01,0x01,0x03,0x03,0x02,0x02,0x01,
    0x01,0x01,0x0b,0x01,0x01,0x01,0x01,0x02,0x02,0x01,0x08,0x01,0x01,0x01,0x01,0x01,0x01,0x02,0x02,0x01,0x01,0x01,0x01,0x01,0x01,0x01,0x0f,0x02,0x02,0x02,0x02,0x02,
    0x02,0x02,0x02,0x02,0x02,0x02
};

unsigned char (*levels)[1] = {&Level1};

I have tried things like:

unsigned char curr_level[70] = levels[0];

but they have not worked.

Thank you!

If 70 is a meaningful number it should be a #define constant of some kind. — tadman
– tadman, Commented Apr 29, 2020 at 0:35

user12986714 · Accepted Answer · 2020-04-29 03:06:38Z

1

WARNING: this answer may be INCORRECT See comments for details.

According to your declaration,

unsigned char (*levels)[1] = {&Level1};

levels is an array of char pointers, and therefore levels[0] is a char pointer.
So depending on what you want to do, you may try one of the followings:

unsigned char *curr_level=levels[0];

or

unsigned char *curr_level;
curr_level=malloc(sizeof(Level1)); /* Don't forget to call free() on it later */
memcpy(curr_level,levels[0],sizeof(Level1));

edited Apr 29, 2020 at 3:06

answered Apr 29, 2020 at 0:30

user12986714

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2 Comments

Eric Postpischil Over a year ago

unsigned char (*levels)[1] declares levels to be a pointer to an array of 1 unsigned char, not to be an array of “char pointers”. levels[0] is an array of 1 unsigned char; it is not a “char pointer”, although, in many uses, it will be converted to a pointer to its element.

user12986714 Over a year ago

@eric Thank you. Is it because of the parentheses(or would there be any difference if it is unsigned char *levels[])? And according to your comment, it is actually incorrect to declare like that (since one is misusing an array as a pointer)?

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