Question

The resize() function makes vector contain the required number of elements. If we require less elements than vector already contain, the last ones will be deleted. If we ask vector to grow, it will enlarge its size and fill the newly created elements with zeroes.

 vector<int> v(20); 
 for(int i = 0; i < 20; i++) { 
     v[i] = i+1; 
    } 
  v.resize(25); 

 for(int i = 20; i < 25; i++) { 
     v[i] = i*2; 
   } 

But if we use push_back() after resize(), it will add elements AFTER the newly allocated size, but not INTO it. In the example above the size of the resulting vector is 25, while if we use push_back() in a second loop, it would be 30.

 vector<int> v(20); 
 for(int i = 0; i < 20; i++) { 
     v[i] = i+1; 
  } 
 v.resize(25); 
  for(int i = 20; i < 25; i++) { 
  v.push_back(i*2);   // Writes to elements with indices [25..30), not [20..25) ! <
 } 

Then where is the advantage of resize() function ? Doesn’t it creates a confusion for indexing and accessing elements from the vector ?

Answer 1

It sounds as though you should be using vector::reserve.

vector::resize is used to initialize the newly created space with a given value (or just the default.) The second parameter to the function is the initialization value to use.