Printing values using an Iterator on a 2d vector
Here is my code:
std::vector< std::vector<std::shared_ptr<int>> > om(2, std::vector<std::shared_ptr<int>>(2));
om[0][0] = std::shared_ptr<int>(std::make_shared<int>(1));
om[0][1] = std::shared_ptr<int>(std::make_shared<int>(2));
om[1][0] = std::shared_ptr<int>(std::make_shared<int>(3)); //init values
om[1][1] = std::shared_ptr<int>(std::make_shared<int>(4));
std::vector<std::shared_ptr<int>>::iterator pd; //inner iterator
std::vector< std::vector<std::shared_ptr<int>> >::iterator px; //outer iterator
for(px = om.begin(); px != om.end(); px++){
for(pd = (*px).begin(); pd != (*px).end(); pd++){
std::cout<< *pd[0] << std::endl;
}
}
Output:
1 2 3 4
I am making a 2d vector using shared_ptr
and my goal is to print the values
My questions:
From other examples I've seen, you only need to dereference the iterator to get the value, but in my case, if I use it std::cout<< *pd << std::endl;
, it will print raw addresses, which is not what I want.
But if I use std::cout<< *pd[0] << std::endl;
then it will print the correct values.
i render it like this:
[ [1][2], [3][4] ]
where px will iterate over 2 blocks and pd will iterate over 4 blocks in total.
Why do I need to [0]
print? or what is undefined?
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Why do I need to put [0] for printing? or what is undefined?
std::vector::iterator
is a random access iterator. This gives operator[]
the same semantics as for a raw pointer. So p[0]
has the effect of de-referencing the iterator, giving you an shared_ptr
lvalue reference . You then remove the link using *
, giving you int
it "points" to.
You could do this too, which would be easier:
std::cout<< **pd << std::endl;
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