Why was a change in vector not reflected in a 2D vector?

Question

I'm new to vectors.

I have two vectors, vector1 and vector2, both have two values each. Now, using these two vectors, I have made a 2 dimensional vector, vector_2d whose value(contents) I want to print. I use the below code and everything works fine.

vector<int> vector1;
vector<int> vector2;

vector1.push_back(10);
vector1.push_back(20);

vector2.push_back(100);
vector2.push_back(200);

vector_2d.push_back(vector1);
vector_2d.push_back(vector2);

cout<<"The elements in vector_2d are: "<<vector_2d.at(0).at(0)<<" "<<vector_2d.at(0).at(1)<<" "<<vector_2d.at(1).at(0)<<" "<<vector_2d.at(1).at(1)<<endl;

Now, I want to replace the first value in vector1 (which is 10) with 1000. I do it by a simple assignment operator:

vector1.at(0) = 1000;

Now, I try to print vector1 and vector_2d again. I get the result that I expected with vector1:

cout<<vector1.at(0)<<endl; //1000

But when I print vector_2d, I get the same result as before. The changes done in vector1 are not being reflected in the 2D vector. Why is this happening?

When you `push_back` into `vector_2d` you are making a copy. It is not the same as the original. — ChrisMM, Nov 28 '19 at 11:09
@ChrisMM Can you elaborate on that please? What does a "copy" mean in this context? Is it that the compiler is creating a similar vector and and using it to `push_back` into the `vector_2d`? If so, then how can I make sure that both these vectors are in sync with each other, and the value changes (if any) are reflected throughout the code? — Aditya Prakash, Nov 28 '19 at 11:14
You can try instead: `vector_2d.push_back(vector()); vector_2d.back().push_back(10); // or vector_2d[0].push_back(10);`. — Scheff's Cat, Nov 28 '19 at 11:16
Internally, it is a bit more complex, but a copy of a vector is as much a copy as a copy of an int is: `int x = 10, int y = x; y = 12;` won't modify `x` either... — Aconcagua, Nov 28 '19 at 11:16
If you *promise* that the two single vectors live at least as long as the 2d vector, you could place them in a vector of pointers or references (you'll need `std::reference_wrapper` then). If you don't keep the promise, you'll end up with dangling pointers or references, so a potentially dangerous approach. — Aconcagua, Nov 28 '19 at 11:18
@Scheff So how exactly is that going to keep the two vectors in sync? If I understand your code correctly, I am pushing value "10" to the first index of `vector_2d`. — Aditya Prakash, Nov 28 '19 at 11:22
Well. The first element of `vector_2d` is of type `vector`. Hence, `push_back()` can be applied. There is no option to _keep the two vectors in sync_. For that, you had to use a `std::vector&>`... (but I wouldn't recommend this.) However, I once wrote an answer which might be of interest: [SO: vector of existing objests](https://stackoverflow.com/a/42039844/7478597) — Scheff's Cat, Nov 28 '19 at 11:24
@Scheff Ahhhh right... sounds a bit like "vector-ception" :P Thanks btw, this helps. — Aditya Prakash, Nov 28 '19 at 11:30
@Scheff `std::vector` is invalid for all `T`. You probably mean `std::vector>>` — Caleth, Nov 28 '19 at 11:45
@Caleth You are correct (and my comment a bit sloppy). Actually, what you proposed is just what I did in the link I provided. — Scheff's Cat, Nov 28 '19 at 11:46

score 1 · Accepted Answer · answered Nov 28 '19 at 13:29

The value in your 2D vector is not changed, because while pushing back 1D vectors

vector_2d.push_back(vector1);
vector_2d.push_back(vector2);

you made copies of each of them. Therefore, changing values in vector1 will not change the values in its copy stored as element 0 of vector_2d. If you would like to change the values in 2D vector, you can do it directly:

vector_2d.at(0).at(0) = 0;

Another possibilities were mentioned in the comments to your question (like having a vector of pointers or references), but I do not recommend those because of possible memory violations (e.g. if your 2D vector will live longer than 1D vectors that are referenced).

Why was a change in vector not reflected in a 2D vector?

1 Answers1