Chapter 12. Dynamic Memory

Exercise 12.1:

How many elements do b1 and b2 have at the end of this code?

StrBlob b1;
{
    StrBlob b2 = {"a", "an", "the"};
    b1 = b2;
    b2.push_back("about");
}

b2 is destroyed, but the elements in b2 must not be destroyed.

so b1 and b2 both have 4 elements.

Exercise 12.2

Exercise 12.3:

Does this class need const versions of push_back and pop_back? If so, add them. If not, why aren’t they needed?

@Mooophy:

You can certainly do this if you want to, but there doesn't seem to be any logical reason. The compiler doesn't complain because this doesn't modify data (which is a pointer) but rather the thing data points to, which is perfectly legal to do with a const pointer. by David Schwartz.

---

Discussion over this exercise on Stack Overflow

Discussion over this exercise more on douban(chinese)

Exercise 12.4:

In our check function we didn’t check whether i was greater than zero. Why is it okay to omit that check?

@Mooophy:

Because the type of i is std::vector<std::string>::size_type which is an unsigned.When any argument less than 0 is passed in, it will convert to a number greater than 0. In short std::vector<std::string>::size_type will ensure it is a positive number or 0.

Exercise 12.5:

We did not make the constructor that takes an initializer_list explicit (7.5.4, p. 296). Discuss the pros and cons of this design choice.

@Mooophy:

keyword explicit prevents automatic conversion from an initializer_list to StrBlob. This design choice would easy to use but hard to debug.

@pezy:

Pros

• The compiler will not use this constructor in an automatic conversion.
• We can realize clearly which class we have used.

Cons

• We always uses the constructor to construct a temporary StrBlob object.
• cannot use the copy form of initialization with an explicit constructor. not easy to use.