Thread: push_back with pointers on STL vectors

I'm going crazy here, I'd really appreciate some help. I have the following piece of code:

Code:

vector<Type*> a_vector;

for (int i=0; i<n; i++) {
	Type* ptr = new Type(this);
	a_vector.push_back(ptr);
	printf("i: %d, ptr: %p\n", i, ptr);
}

printf("size: %d, a_vector[0]: %p\n", a_vector.size(), a_vector[0]);

And I get this result:

Code:

i: 0, ptr: 0x80c8530
i: 1, ptr: 0x80c8648
i: 2, ptr: 0x80c8738
i: 3, ptr: 0x80c8828

size: 4, a_vector[0]: (nil)

I don't get it: I print ptr before doing push_back and it isn't nil, the final size of a_vector is 4, which shows it has indeed inserted all the pointers, but then the first one (and the others too, actually) is nil. How is this possible? Thanks a lot.

Ok, don't mind me. I'm stupid. *sigh*. In case anyone's wondering, the problem was that I resized a_vector and then added elements at the back. In my defense I must say that it's been a looong loooooong time since I last used the STL (or C++, for that matter), and I had a really big test this morning, so my brain isn't working that good.

The problem is solved now but I feel really stupid

why are you using printf?

I was just saying, using the C++ STL while using a C standard output format seems counter intuitive

that does seem counter intuitive, i agree. too bad 90% of the c++ tutorials i search through support it. the craziness. i prefer cout.

Use std::cout instead.

I know, I know, I'm not actually going to print anything in the final program, I was just checking on the values of my variables. I like cout way better than printf myself but it's been so long since I programmed in C++ and I'm so used to C my mind plays tricks on me sometimes. Thanks for the suggestion anyway.

And you can always do something like
[code]vector.push_back(new type value) or something. Instead of having an intermediate value to push back. Don't forget to traverse the vector and delete any values when not needed, since just deleting a record will not do that for you.