Programming

There's nothing weird about this...as Corona688 noted you cant assign a pointer the address of a constant...your compiler is broken so get a new one.

edit: nevermind...your compiler's parser is more off than my mental parser, lol.

Operator precedence: the first expression returns the address of x which is then incremented. The second one attempts to increment the address of x. I think.

What was your intent?

DonDoerner: What is the address you're supposed to get from &(x-1) ?

There isn't one of course, since a mathematical expression isn't a variable. Post-increment and pre-increment are in the same category.

So, if I recall correctly, '&' and '++' have the same level of precedence, and are evaluated left-to-right. If this is the case, the first expression increments x, then takes the address of it. Though likely meaningless, this is legal from the code snippet we are seeing.

The second expression takes the address of x and increments it, which the compiler can recognize as nonsense.

But again, I have to ask: what was the intent, what did the author of this question intend?

&x++ is nonsense either way.

If it takes the & first and converts X into an address, it cannot do ++ on it because ++ requires an lvalue, not an expression.

If it takes the ++ first, then tries to take the address of it, it cannot, because x++ is an expression, not an lvalue.

Only the OP can tell you what he wanted, and he never came back...

The result is undefined.