Difference between i = ++i and ++i [duplicate]

问题

Possible Duplicate:
Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)

What is the difference between i = ++i; and ++i; where i is an integer with value 10?

According to me both do the same job of incrementing i i.e after completion of both the expressions i =11.

回答1:

i = ++i; invokes Undefined Behaviour whereas ++i; does not.

C++03 [Section 5/4] says Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression.

In i = ++i i is being modified twice[pre-increment and assignment] without any intervening sequence point so the behaviour is Undefined in C as well as in C++.

However i = ++i is well defined in C++0x :)

回答2:

Writing i = ++i; writes to variable i twice (one for the increment, one for the assignment) without a sequence point between the two. This, according to the C language standard causes undefined behavior.

This means the compiler is free to implement i = ++i as identical to i = i + 1, as i = i + 2 (this actually makes sense in certain pipeline- and cache-related circumstances), or as format C:\ (silly, but technically allowed by the standard).

回答3:

i = ++i will often, but not necessarily, give the result of

i = i;

i +1;

which gives i = 10

As pointed out by the comments, this is undefined behaviour and should never be relied on

while ++i will ALWAYS give

i = i+1;

which gives i = 11;

And is therefore the correct way of doing it

回答4:

If i is of scalar type, then i = ++i is UB, and ++i is equivalent to i+=1. if i is of class type and there's an operator++ overloaded for that class then i = ++i is equivalent to i.operator=(operator++(i)), which is NOT UB, and ++i just executes the ++ operator, with whichever semantics you put in it.