vector.erase() question

[H00G0]

Hello!

I have a simple question about a vector operation.

Let's say we have this:

vector<int> numbers;

numbers.push_back(0);
numbers.push_back(10);
numbers.push_back(20);
numbers.push_back(30);
numbers.push_back(40);

If I erase one of the number like so:

numbers.erase(2);

Does it set number[2] equal to 0 without changing the vector size,
or does it remove the int held in number[2] and decrement the vector's size by one?

Ultimately what I want to do is have a vector in which I would erase its content one by one,
and I want to check if it's empty with if (numbers.empty()) so if its size is 0.

Thank you in advance!

[Repeater]

Before: { 0 10 20 30 40 } , size 5

numbers.erase(numbers.begin() + 2);

After: { 0 10 30 40 } , size 4

[H00G0]

Okay, thank you Repeater this is exactly what I wanted to know!

Have a good day!

[Chervil]

If I erase one of the number like so:

    numbers.erase(2);

Then you will get a compiler error.

The erase function takes an iterator as a parameter. Iterators are a topic in their own right, but you can think of it as an object which behaves like a pointer.

See the reference page for erase():
http://www.cplusplus.com/reference/vector/vector/erase/

One possible syntax for your example would be:

    // erase one of the numbers    
    numbers.erase(numbers.begin() + 2);

and it reduces the array size by 1.

(see the reference page for begin() as well).
http://www.cplusplus.com/reference/vector/vector/begin/


Example code below:

#include <iostream>
#include <iomanip>
#include <vector>

using namespace std;

void print(const vector<int> & v)
{
    cout << "size = " << v.size() << '\n';
    for (int n : v)
        cout << setw(4) << n;
    cout << '\n';        
}

int main()
{
    vector<int> numbers;
    
    numbers.push_back(0);
    numbers.push_back(10);
    numbers.push_back(20);
    numbers.push_back(30);
    numbers.push_back(40);    
    
    // Now print it out
    print(numbers);
    

    // erase one of the numbers    
    numbers.erase(numbers.begin() + 2);

    print(numbers);	    
}

Output:

size = 5
   0  10  20  30  40
size = 4
   0  10  30  40

[H00G0]

Thank you Chervil for the detailed explanation and the sources. I was aware of the .size and .erase functions however I didn't know how .erase interacted with the vector size.

So to triple check if you do this:

numbers.erase(number.begin(), number.end());

numbers.size() should be equal to zero
and the content of numbers should be {}.

Is that correct?

[Chervil]

Is that correct?

Yes. (after correcting minor typing errors).

It's easy enough to check using the online compiler. If you look at my post above, to the top right of the code there is a gear icon. Click that and it takes you to a page where you can edit, compile and run the code:

http://cpp.sh/73km5

[H00G0]

Thank you Chervil I will keep that in mind for later posts !