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.
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.
array[3] certainly is undefined behavior, but the initialization of array itself is well-defined such that the initialization of array above is equivalent to char array[] = { '1', '2', '3' };. Room for the null terminator is not requisite. On the other hand, char array[3] = "1234"; is undefined since "No initializer shall attempt to provide a value for an object not contained within the entity being initialized" (C11 draft N1570 6.7.9p2), meaning that array is too small to contain the 4 elements { '1', '2', '3', '4' }. 6.7.9p32 of N1570 expresses the intended behavior.
'\0' is not supplied?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.
