C++, comparing pointers.

[cgroza]

Hello everyone. I started to do a limited container just to test my skills after finishing the chapter about iterators in the book "The C++ programming language".

So my "container" basically uses a vector(by default, can be specified as a template argument) to hold its elements. I made this whole thing just to try to implement iterators for it.

Here is the container class code:


#include <iostream>
#include <vector>
using namespace std;

template <class T, class C> class Vec {
private:
typedef T val_type;

C vec;
struct ends {
val_type* first;
val_type* last;
};

ends extrem;

public:
template<class V = val_type>
class iterator {
private:
V* it;
public:
iterator(val_type* a):it(a) {
}

void operator=(val_type* a) {
it=a;
}

void operator++() {
++it;
}
val_type& operator*()const {
return *it;
}
bool operator<(val_type* y) {
return it < y;
}
};

Vec(const int nr):vec(nr) {
extrem.first = &vec[0];
extrem.last = &vec[-1];

}

val_type* end() const {
return extrem.last;
}

val_type* begin() const {
return extrem.first;
}

void push_back(const val_type& a) {
vec.push_back(a);
extrem.first = &vec[0];
extrem.first = &vec[-1];
}

val_type& operator[](int x) {
return vec.at(x);
}

int size() {
return vec.size();
}

};And here is my main:


int main()
{
Vec<int, vector<int> > test(10);
Vec<int, vector<int > >::iterator<int> p = test.begin();


for(int j = 0 ; j < test.size() ; j++) {
test[j] = j;
}

int i = 0;
while (i<test.size()) {
cout<<*p;
++p;
i++;
}
}With this main function, the thing prints out all the elements using my iterator.
The problem is that I want to do the loop with while(p<test.end()){...} like in the book, but the < operator returns something bad and the loop goes crazy and segfaults.
This is the < operator:


bool operator<(val_type* y) {
return it < y;
}Any ideas why? I am I doing something wrong?

[GeneralZod]

This seems highly dubious:





extrem.last = &vec[-1];

[cgroza]

This seems highly dubious:
More explanation please?

[PaulM1985]

You seem to be saying the last position is the one before the first.

Paul

[cgroza]

You seem to be saying the last position is the one before the first.

Paul
OK, I think that I was thinking python. In python -1 is the last position for an array. :D
EDIT: Great, it works. Now lets do a google on python mistakes in C++ :D.

[Vox754]

This seems highly dubious:



>>> a = ["first", 1, 2, 3, "last"]
>>> print(a)
['first', 1, 2, 3, 'last']
>>> print(a[0])
first
>>> print(a[-1])
last


It works in Python!

I wonder why it doesn't work in C++?

[cgroza]

>>> a = ["first", 1, 2, 3, "last"]
>>> print(a)
['first', 1, 2, 3, 'last']
>>> print(a[0])
first
>>> print(a[-1])
last
It works in Python!

I wonder why it doesn't work in C++?

Different languages, maybe it is because of the vector, let me try for a built in array.
EDIT: It does not, just spits some random data.

[Vox754]

Different languages, maybe it is because of the vector, let me try for a built in array.
EDIT: It does not, just spits some random data.

You were supposed to read my witty remark before realizing your mistake. Otherwise my post makes no sense.