Set iterator for loop decrementer position

[boogerlad]

code:

std::set <int> faster;     std::set <int>::iterator itr;     std::set <int>::reverse_iterator rit;     faster.insert(15);     faster.insert(22);     faster.insert(9001);     faster.insert(158);     faster.insert(220);     faster.insert(90501);     faster.insert(1115);     faster.insert(232);     faster.insert(9001);     for (itr = faster.end(); itr != faster.begin(); --itr)     {         std::cout << *itr << std::endl;     }     for (rit = faster.rbegin(); rit != faster.rend(); ++rit)     {         std::cout << *rit << std::endl;     }

Those two for loops should print exactly the same, but it doesn't. However, if I move the --itr like so:

code:

for (itr = faster.end(); itr != faster.begin();)     {         --itr;         std::cout << *itr << std::endl;     }

then it works out fine. Can anyone explain why this is so? I would've thought that having the decrementer in the for loop statement and the first line would make no difference in execution. Btw, I know you can use a reverse iterator, but I'm just curious as to why this happens.

[Tony]

code:

for( A ; B ; C )    D

is functionally equivalent to

code:

A while(B) {   D   C }

That should be enough for you to figure out why moving that last statement into the loop is different.

[boogerlad]

Thanks so much! It all makes sense now!

[boogerlad]

wait a minute, I understand the differences in the placement of the decrementer, but now I have another question. If the decrementer returns an object (in this case, an integer), then why isn't the first element skipped when printing it out?

[Tony]

which code example are you referring to?

[boogerlad]

code:

std::set <int> faster;     std::set <int>::iterator itr;     faster.insert(15);     faster.insert(22);     faster.insert(9001);     faster.insert(158);     faster.insert(220);     faster.insert(90501);     faster.insert(1115);     faster.insert(232);     faster.insert(9001);     for (itr = faster.end(); itr != faster.begin(); )     {         --itr;         std::cout << *itr << std::endl;     }

This works perfectly, but wouldn't the first thing that get's printed not be the last item in the set because it's getting decremented and the iterator returns an integer?

[Tony]

you are wrong on two accounts

1) iterator is not an integer, it's of type iterator
2) #end http://www.cplusplus.com/reference/stl/set/end/

Quote:

Returns an iterator referring to the past-the-end element in the set container.

[boogerlad]

hmmm, thanks. I should probably read the stl api a bit more thoroughly before I make my self look like an idiot, thanks!