In C language, an array cannot be copied to another array directly by assignment operator.
int arr1[]={1,3,2}; int arr2[]={0}; arr2=arr1;//not possible Also, we cannot assign values to an array that is already defined, if I am not wrong...
int a[3]; a[]={1,3,2}; //this is not possible In the code above, a[] and {1,3,2} act as two different arrays, and an assignment operator is used between them. So, is this following the same case mentioned at the first?
Please clarify. Thanks.
is this following the same case mentioned at the first?
No, they are different.
In the first case, what you try to do is array assignment, which is not possible in C directly. The language grammar states an array name to be a non-modifiable lvalue, so
arr1 = arr2; is invalid. The relevant excerpt from the C11 standard draft (§6.3.2.1):
A modifiable lvalue is an lvalue that does not have array type, ...
In the second case
int a[3]; a[]={1,3,2}; you are trying to assign the initializer list {1, 2, 3} to an invalid sub-expression a[] and hence is illegal. What is assigned to it is immaterial; a[] = 1 is invaild too.
It's not quite the same.
{1,2,3} is an initializer list. The compiler sees it and hard-codes its values into the array. Assigning an initializer-list to an array that already has a definition (int array[3]; array={1,2,3};) is a syntax error (clang says expected expression because it's expecting a closing square bracket before an assignment operator as in int array[]={1,2,3};).
Assigning one array to another is impossible in your case because array type 'int [3]' is not assignable. I'd call it something like logic error (the programmer just didn't know he can't do it, but the syntax is valid).
The other posts and comments are correct in that you cannot assign a bare array after initialization.
However, you can assign to a struct that contains such an array, even after initialization, using a compound literal,
typedef struct { int v[3]; } A; A a = {{1,2,3}}; // Valid. a = (A){{4,5,6}}; // Also valid. 
arr1andarr2are 1D arrays), but a single element. This is very well possible. Please clarify your question. From a wrong prerequisite, everything can be deduced.memsetormemcpywill help you to achieve the same thing though:memcpy(arr1, arr2, sizeof arr1/sizeof *arr1);or something