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by poepie 1298 days ago
In the first example a & b are variables, which can be assigned to each other. In the second a & b are pointers, but b is fixed, so you can not assign a value to it.
1 comments

They’re not pointers. sizeof a == 4*sizeof(int), not sizeof(int*).
They're pointers, just weird ones. The compiler knows it's an array, so it gives the result of the actual amount of space it takes up. If you passed it into a function, and used the sizeof operator in the function, it'd give `sizeof(int *)`. Because sizeof is a compile-time operation, so the compiler still knows that info for your example.
That jest means it decays into a pointer after being passed as a function argument. In the example given however it’s not a pointer. Just like it wouldn’t be inside a struct.