2 &x = 20; // this doesn't work
3 * (&x) = 20; // this does work
Why does line 2 &x not work but line 3 does? Because &x returns a pointer, a number representing a memory address. This is an important distinction. A pointer doesn’t hold a memory address, it holds a number that represents a memory address.
=======
No, that is not why. Note that the following does work:
int * x = 0;
and the following works, though typically yields a warning:
int * x = 20;
Line 2 fails because & doesn't give back an l-value.
No comments yet.