C++ cout inside loop for printing vector of vectors fails

Question

template<typename s>
void vecprint2d(const s& vec){
    cout<<"{"<<endl;
    for(int x = 0; x < vec.size(); x++){
        cout<<"{";
        for(int y = 0; y < vec[x].size() - 1;y++){
            cout << vec[x][y]<<", ";
        }
        cout<<vec[x][vec[x].size() - 1]<<"}"<<endl;
    }
    cout<<"}"<<endl;
}


int main(){
vector<vector<int>> vec = {{1,2,3},{},{4,5,6}};
vecprint2d(vec);
return 0;
}

in my attempt at a function for printing a vector of vectors, why does cout inside the inner loop cause problems, or is the problem elsewhere?

the output right now looks like:

{

{

How are you calling this function? Please add the calling code snippet also. — Azeem, Apr 10 '20 at 16:20
[cannot reproduce](https://wandbox.org/permlink/ojjdbGotavyZnfqt) — 463035818_is_not_an_ai, Apr 10 '20 at 16:26
is this really the code you are running to get that output? (i know it isnt, because there is a missing `}` but I dont know if this is just a typo here or if you typed code here that it completely different from your real code) — 463035818_is_not_an_ai, Apr 10 '20 at 16:29
your code will access index -1 if one of the inner vectors is empty — 463035818_is_not_an_ai, Apr 10 '20 at 16:30
Other than the missing `}` pointed out by @idclev463035818 and a missing `int` before `main`, your posted code gives me the output you/I would expect (VS2019/MSVC or clang-cl). — Adrian Mole, Apr 10 '20 at 16:31
well if that is the input it is understandable why the function does not work properly.... — 463035818_is_not_an_ai, Apr 10 '20 at 16:31
i left out an empty set of {} in vec in the example, does that make a differnece? — 2wings, Apr 10 '20 at 16:32
the difference is that this code is invalid while the one before was ok and there was no way to explain the output you got. Details do matter! — 463035818_is_not_an_ai, Apr 10 '20 at 16:33
[Idioms-for-for-each-except-the-last](https://stackoverflow.com/questions/35372784/idioms-for-for-each-except-the-last-or-between-each-consecutive-pair-of-el/35373017#35373017) might interest you. — Jarod42, Apr 10 '20 at 17:15

score 3 · Accepted Answer · answered Apr 10 '20 at 16:33

If the inner vector's size is 0, size() - 1 will overflow and it will loop forever and/or crash. Could that be what is happening on your raspberry pi?

To avoid this, handle 0-sized vectors as well.

For example like this:

template<typename s>
void vecprint2d(const s& vec) {
    cout << "{" << endl;
    for (auto const& row : vec) {
        cout << "{";
        int i = 0;
        for (auto const& val : row) {
            if (i++)
                cout << ", ";
            cout << val;
        }
        cout << "}" << endl;
    }
    cout << "}" << endl;
}

Adrian Mole · Answer 2 · 2020-04-10T16:50:20.763

The size() function of an empty vector will return an (unsigned) value of zero and, in your inner loop, you test y against vec[x].size() - 1. This will give a value that has 'underflowed' and thus have the maximum value that a size_t variable can hold, so the y loop will run a very large number of times! However, it will likely fail on the first loop, because trying to access any element of an empty vector is undefined behaviour.

To fix this, enclose your inner loop in an "is it empty" if block (in fact, you should do this for both loops). Here's a possible solution:

template<typename s>
void vecprint2d(const s& vec)
{
    cout << "{" << endl;;
    if (!vec.empty()) for (size_t x = 0; x < vec.size(); x++) {
        cout << "{";
        if (!vec[x].empty()) {
            for (size_t y = 0; y < vec[x].size() - 1; y++) {
                cout << vec[x][y] << ", ";
            }
            cout << vec[x][vec[x].size() - 1];
        }
        cout << "}" << endl;
    }
    cout << "}" << endl;
}

Feel free to ask for further clarification and/or explanation.

C++ cout inside loop for printing vector of vectors fails

2 Answers2