a[14]={0}?

[sybariticak47]

Hi! I've seen in some books notation like :

a[14]={0};
char a[28]={'#'}

What does this do? Thanks!

[joeprogrammer]

That does not work because you are redifining "a"; try it; you will end up with a compiler error. I have no idea why they would do that.

Unless those are 2 different lines of code (ie, in a different program). In that case, it probably fills the first element of the array with the charecter.

[sybariticak47]

yeah they are two separate codes. So it only fill the first element?

[laserlight]

So it only fill the first element?

Not quite. In the case of:

int a[14] = {0};

the array is zero initialised, i.e. all 14 elements will be set to 0.
For the string:

char b[28] = {'#'};

the first character is set to '#', the rest are left as null characters.

[sybariticak47]

Thanks for the help

[7stud]

Hi,

Just by way of further explanation: if you use an initializer list when you declare an array, i.e.:

int nums[3] = {1, 2};

and your initializer list does not contain the same number of elements as the size of the array, the remaining elements in the array will be automatically initialized to 0. 0 will be converted to the appropriate type for the array, e.g. a 0 for a char array is the '\0' character. So you can initialize a whole array to 0 doing this:

int char[20] = {0};

The first element is initialized to 0 or a '\0' character, and since the initializer list doesn't contain values for the rest of the array, the remaining elements are automatically initialized to '\0'. Now, if you do this:

char a[28]={'#'};

That assigns '#' to the first element in the array, and since the initializer list doesn't contain values for the rest of the array, the remaining elements are automatically initialized to '\0'.