++i or i++ in for loops ?? [duplicate]

问题

Possible Duplicate:
Is there a performance difference between i++ and ++i in C++?

Is there a reason some programmers write ++i in a normal for loop instead of writing i++?

回答1:

For integers, there is no difference between pre- and post-increment.

If i is an object of a non-trivial class, then ++i is generally preferred, because the object is modified and then evaluated, whereas i++ modifies after evaluation, so requires a copy to be made.

回答2:

++i is slightly more efficient due to its semantics:

++i;  // Fetch i, increment it, and return it
i++;  // Fetch i, copy it, increment i, return copy

For int-like indices, the efficiency gain is minimal (if any). For iterators and other heavier-weight objects, avoiding that copy can be a real win (particularly if the loop body doesn't contain much work).

As an example, consider the following loop using a theoretical BigInteger class providing arbitrary precision integers (and thus some sort of vector-like internals):

std::vector<BigInteger> vec;
for (BigInteger i = 0; i < 99999999L; i++) {
  vec.push_back(i);
}

That i++ operation includes copy construction (i.e. operator new, digit-by-digit copy) and destruction (operator delete) for a loop that won't do anything more than essentially make one more copy of the index object. Essentially you've doubled the work to be done (and increased memory fragmentation most likely) by simply using the postfix increment where prefix would have been sufficient.

回答3:

++i is a pre-increment; i++ is post-increment.
The downside of post-increment is that it generates an extra value; it returns a copy of the old value while modifying i. Thus, you should avoid it when possible.

回答4:

With integers, it's preference.

If the loop variable is a class/object, it can make a difference (only profiling can tell you if it's a significant difference), because the post-increment version requires that you create a copy of that object that gets discarded.

If creating that copy is an expensive operation, you're paying that expense once for every time you go through the loop, for no reason at all.

If you get into the habit of always using ++i in for loops, you don't need to stop and think about whether what you're doing in this particular situation makes sense. You just always are.

回答5:

There is a reason for this: performance. i++ generates a copy, and that's a waste if you immediately discard it. Granted, the compiler can optimize away this copy if i is a primitive, but it can't if it isn't. See this question.

回答6:

No compiler worth its weight in salt will run differently between

for(int i=0; i<10; i++)

and

for(int i=0;i<10;++i)

++i and i++ have the same cost. The only thing that differs is that the return value of ++i is i+1 whereas the return value of i++ is i.

So for those prefering ++i, there's probably no valid justification, just personal preference.

EDIT: This is wrong for classes, as said in about every other post. i++ will generate a copy if i is a class.

回答7:

As others have already noted, pre-increment is usually faster than post-increment for user-defined types. To understand why this is so, look at the typical code pattern to implement both operators:

Foo& operator++()
{
    some_member.increase();
    return *this;
}

Foo operator++(int dummy_parameter_indicating_postfix)
{
    Foo copy(*this);
    ++(*this);
    return copy;
}

As you can see, the prefix version simply modifies the object and returns it by reference.

The postfix version, on the other hand, must make a copy before the actual increment is performed, and then that copy is copied back to the caller by value. It is obvious from the source code that the postfix version must do more work, because it includes a call to the prefix version: ++(*this);

For built-in types, it does not make any difference as long as you discard the value, i.e. as long as you do not embed ++i or i++ in a larger expression such as a = ++i or b = i++.