1.2.1.7 Sizes

• For container, Call empty() instead of of checking size() against zero. it’s very easy to understand it.
• for vector, size(), capacity() and max_size() can be seen below: capacity is equal or bigger than size. So if you want to avoid allocate the unnecessary space. LLVM give a SmallVecotr<type, N> example, you can use N to specify a smaller vector to avoid waste.
  std::vector<int> myvector; 
  for (int i=0; i<100; i++) myvector.push_back(i); 
   
  std::cout << (int) myvector.size() << ’\n’; // 100 
  std::cout  << (int) myvector.capacity() << ’\n’;  //128 
  std::cout << (int) myvector.max_size() << ’\n’;  //1073741823
• Four confused conceptions:
  capacity() //how many CAN hold 
  size()   //how many are in NOW 
  resize(n) // forces the container to change to n, 
  //if n <size(),  object in the end will be lost 
  // if n>size(),  default ctor of element will be called. 
  // after resize(n), size will return n. 
  // but resize don’t change capacity 
   
  reserve(n) //cause the container’s capacity() to at lease n.
• You want to avoid allocation, 1), if you know the number, then use reserve, 2) if you don’t know the number, you can reserve maximum space, then you can trim off any excess capaciy. you can use shrink_to_fit() function too. Just remember string(s).swap(s)
  vector<class> v1; 
  v1.reserve(1000); v1.size(); // only 5 
   
  vector<class>(v1).swap(v1); 
  //1) vector<class>(v1) copy ctor create a temp vector 
  //2) temp just copy real object, so it’s capacity (maybe 8)is small. 
  //3) temp.swap(v1) then temp has 1000 space,  v1 capacity is 8 now 
  //4) in the end of statment, temp destory. 
  string(s).swap(s) // the same idea behind.
• Worthwhile note about vector: It will avoid allocate new space many times.
  std::vector<int> v 
  for(){ ... 
   v.clear 
  }  //good smell. 
   
  for(){vector<int> v ...} //bad smell