Why does int* ptr_arr_int = {1,2}; not work in C/C++?

"amruth" is a const char[7] type in C++, and a char[7] type in C (although the behaviour on attempting to modify the string is undefined).

This can decay to a const char* or char* type respectively in some circumstances, such as yours.

Although an int[2] will similarly decay to an int* in some circumstances, {1, 2} is neither an int[2] nor a const int[2] type; rather it is a brace-initialiser.

As said, the string is an const char[7] array and while that can decay to a char*, {1, 2} cannot decay to int* (it's an brace initializer list, similarly: std::initializer_list). But note there is also the array [] declaration which allows you to automatically declare arrays. Combine that with the list-initialization since C++11 and you are able to initialize both via []:

int ptr_arr_int[] = { 1,2 }; //OK
char ptr_arr_char[] = "amruth"; //OK

A string by itself already implies a storage class -- it's a static (+effectively const) char array declared using special syntax.

In contrast to that, it's not clear how {1, 2} in int *ptr_arr_int = {1, 2} should be stored. Should it be static or auto?

If you want it auto when in a local scope or static when in file scope, then in C>=99 you can explicitly do int *ptr_arr_int = &(int[]){1,2}[0]; (the &[0] is optional).

You could conceivably make this implicit, but how far would you take this?int ****x = {1,2}; initialized by creating an array, a pointer to it, a pointer to that etc., could result int quite a bit of code/data and hiding a lot of code/data under a simple construct isn't quite the C way. It would also not be clear when the brace syntax should be initializing the ultimate target basic type or some of the intermediate pointer(s).