terminate called after throwing an instance of 'std::out_of_range'

Question

Why does this happen my program says that it has no errors but then when I run it I get terminate called after throwing an instance of \'std::out_of_range\' what(): vector:_M_ra

Answer 1

It looks to me as if this is due to a typo, and you should use the variable 'j' in the second loop. After the first loop,

for (i = 0; i < 52; i++)
{
    nums.push_back(i);
}

the variable 'i' contains the value 52, so it sounds expected that calling nums.at(i) would throw a std::out_of_range, since nums only contains 52 values, starting at index 0.

for(int j = 0; j < 52; j++)
{
    cout << nums.at(i) << "\n";
}

Fix it by replacing the argument of at() with 'j', which I assume was the original intent:

for(int j = 0; j < 52; j++)
{
    cout << nums.at(j) << "\n";
}

Answer 2

You access elements in deck here:

num1 = deck.at(pos1);
num2 = deck.at(pos2);

but it is empty. You have to fill it at some point before making those calls. You can check if a vector is empty with the std::vector::empty method, and get it's size with std::vector::size. See this std::vector reference for more information on those two methods.