以下两种情况初始化向量v(n)有什么区别

Question

I have following piece of code,

#include <iostream>
#include <vector>
using namespace std;

int main()
{
    int input,n;
    cin >> n;
    vector<int> v(n);
    for(int i=0;i<n;i++){
       cin>>input;
       v.push_back(input);
    }   
   for(int i=0; i<v.size();i++){
       cout << v[i] << endl;
   }
   cout << v.size() << endl;
   return 0;
}

and

#include <iostream>
#include <vector>
using namespace std;

int main()
{
    int input,n;
    vector<int> v(n);
    cin >> n;
    for(int i=0;i<n;i++){
       cin>>input;
       v.push_back(input);
    }

   for(int i=0; i<v.size();i++){
       cout << v[i] << endl;
   }
   cout << v.size() << endl;
   return 0;
}

for the following input, n=5, 1 2 3 4 5 , first program gives the output

0
0
0
0
0
1
2
3
4
5
10

and the second program gives the result

1
2
3
4
5
5

I donot understand why vector is initializing 0's in the first doubling the size but not in the second

Answer 1

Actually, neither of those are correct for what you want (I'm assuming here you want the five elements you input and nothing more - that may be an invalid assumption but it seems like a good bet to me).

---

In your first one, you have (comments added to explain what's happening):

int input,n;       // n has arbitrary value.
cin >> n;          // overwritten with 5.
vector<int> v(n);  // initial vector has five zeros.

This means your vector will be created with five (assuming you input that for n ) default-constructed entries. Then you go and insert five more, which is why you get five zeros followed by your actual data.

---

Your second one has:

int input,n;       // n has arbitrary value.
vector<int> v(n);  // initial vector of that arbitrary value (probably).
cin >> n;          // overwritten with five but it's too late anyway.

This is actually undefined behaviour since you have no idea what n will be when you create a vector of that size. In your case, it appears to be zero but that's just a lucky coincidence.

Note the use of the word "probably" above. While it's likely that the vector will be created with some arbitrary number of entries, undefined behaviour means exactly that - you should not rely on it at all. By rights, it could quite easily delete all your files while playing derisive_laughter.wav through your sound system:-)

The most likely case however will be one of:

• it'll work as you thought, because n is zero;
• it'll have some arbitrary number of zeros at the start, similar to your first code snippet;
• it'll cause an out-of-memory exception because n is ridiculously large; or
• it'll cause an exception because n is negative.

---

In any case, you should generally either pre-allocate and just set the already-existing values:

vector<int> v(2);
v[0] = 7;
v[1] = 42;

or not pre-allocate, using push_back to expand the vector on demand:

vector<int> v;
v.push_back(7);
v.push_back(42);

Answer 2

Hope, one can write the 1st case of writing program in the question in the following way to get correct result

1
2
3
4
5
5 

for the input given as in the question

#include <iostream>
#include <vector>
using namespace std;

int main()
{
   int n;
   cin>>n;
   vector<int> v(n);

   for(int i=0;i<n;i++){
       cin>>v.at(i);           
   }
   for(int i=0;i<v.size();i++){
       cout << v[i] << endl;
   }
   cout << v.size() << endl;
   return 0;
}