static char array declaration in C

Question

Can someone explain to me why int array[3] = {1,2,3} work but char array[3] = "123" doesn't work?

It printed out " 123( ( " instead of " 123 ".

It say that for char array need another space for null character but doesn't array start from 0 so char array[3] is enough since it actually 4 space. Unless char array actually require 2 space one for null and one for a special character.

Answer 1

int array[3] = {1,2,3}

allocates an array that can hold at most 3 integers and you can access any of them using the following ways:

array[0]
array[1]
array[2]

Here:

char array[3] = "123"

"123" consist of 4 bytes. The characters in the string literal ends with a NUL-terminator. But you allocate 3 bytes and can atmost hold 2 characters (+1 for the NUL-terminator). So, when you initialize the array with it, the '\0' is not written as there is no space.

When you print this using %s in a printf, it invokes Undefined behavior as %s prints everything until a NUL-terminator. So, the printf goes on reading values from invalid memory locations until a '\0'. Anything can happen when you do this. You might see random garbage getting outputted, crashes, segmentation faults etc. Basically, you should never rely on this behavior, even if it seemed to have "worked" as expected.

but doesn't array start from 0

Yes

so char array[3] is enough since it actually 4 space

Wrong. The valid indices for char array[3] are

array[0]
array[1]
array[2]

accessing array[3] is wrong and it invokes Undefined Behavior.

---

Fix the issue by using

char array[4] = "123";

or better:

char array[] = "123";

The empty brackets tell the compiler to determine the size of the array.

Answer 2

When declaring an array in c, char array[3] will yield 3 chars not 4 like you suggested in your question. You do need space for a null character at the end of a string so you need char array[4] in this case.