4 ms·
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 t
by poepie 4y 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.
- yakubin 4y agoThey’re not pointers. sizeof a == 4*sizeof(int), not sizeof(int*).
- chipsa 4y agoThey'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.
- yakubin 4y agoThat 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.