++*f = ++*g;
What will happen in the above code?

Question

++*f = ++*g;
What will happen in the above code?

Bryan · Accepted Answer

Dearest Ludwig,

Um... No. Did you actually get this compile error or are you speculating? 

Here is a real answer:
Prefix and dereference operations have the same precedence and associate right-to-left.
So, the contents of "g" are incremented before being set to the value of "f" which are also incremented. If we are talking ints, say *f=2 and *g=4, then dereference happens, values are incremented, then the incremented value of *g is assigned to *f. The prefix on the left hand side is overwritten by the assignment and *f=*g=5.

Ludwig · Answer

That causes compile error ("Lvalue required as left operand of assignment").

This is because statement ++*g is identical to ++(*g), which means we get value pointed by g, increment it, and assign it to something else.  But the left statement ++*f is also rvalue (not lvalue), hence error.

But:
*f++ = *g++;

or:

++f;
*f = ++*g;

or:
*++f  = ++*g;

is valid.

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