5-x-chapter-5-summary-and-quiz-processed

测试时间

问题 #1

Why are named constants often a better choice than literal constants?

答案

Using literal constants (a.k.a. magic numbers) in your program makes your program harder to understand and harder to modify. Symbolic constants help document what the numbers actually represent, and changing a symbolic constant at its declaration changes the value everywhere it is used.

Why are const/constexpr variables usually a better choice than defined symbolic constants?

答案

define constants do not show up in the debugger and are more likely to have naming conflicts.

问题 #2

Find 3 issues in the following code:

#include <cstdint> // for std::uint8_t
#include <iostream>
 
int main()
{
  std::cout << "How old are you?\n";
 
  std::uint8_t age{};
  std::cin >> age;
 
  std::cout << "Allowed to drive a car in Texas: ";
 
  if (age >= 16)
      std::cout << "Yes";
  else
      std::cout << "No";
 
  std::cout << '.\n';
 
  return 0;
}

Sample desired output:

How old are you?
6
Allowed to drive a car in Texas: No
 

How old are you?
19
Allowed to drive a car in Texas: Yes
 

答案

1. On line 8, age is defined as a std::uint8_t. Because std::uint8_t is typically defined as a char type, using it here will cause the program to behave as if we’re inputting and outputting a char value rather than a numeric value. For example, if the user entered their age as “18”, only the character '1' would be extracted. Because 1 has ASCII value 49, the user would be treated as if they were 49 years old. A regular int should be used to store the age, as the age doesn’t require a specific minimum integer width. We can also remove #include <cstdint>.
2. On line 13, we use the magic number 16. Although the meaning of 16 is clear from the context it is used in, using a constexpr variable with the value 16 should be preferred instead.
3. On line 18, '.\n' is a multicharacter literal that will print the wrong value. It should be double-quoted (".\n").

问题 #3

What are the primary differences between std::string and std::string_view?

What can go wrong when using a std::string_view?

答案

std::string provides a modifiable string. It is expensive to initialize and copy.

std::string_view provide a read-only view of a string that exists elsewhere. It is inexpensive to initialize and copy. std::string_view can be dangerous when the string being viewed is destroyed before the std::string_view that is viewing it.

问题 #4

Write a program that asks for the name and age of two people, then prints which person is older.

Here is the sample output from one run of the program:

Enter the name of person #1: John Bacon
Enter the age of John Bacon: 37
Enter the name of person #2: David Jenkins
Enter the age of David Jenkins: 44
David Jenkins (age 44) is older than John Bacon (age 37).
 

提示

Hint: Input the person’s name using std::getline()

答案

#include <iostream>
#include <string>
#include <string_view>
 
std::string getName(int num)
{
    std::cout << "Enter the name of person #" << num << ": ";
    std::string name{};
    std::getline(std::cin >> std::ws, name); // read a full line of text into name
 
    return name;
}
 
int getAge(std::string_view sv)
{
    std::cout << "Enter the age of " << sv << ": ";
    int age{};
    std::cin >> age;
 
    return age;
}
 
void printOlder(std::string_view name1, int age1, std::string_view name2, int age2)
{
    if (age1 > age2)
        std::cout << name1 << " (age " << age1 <<") is older than " << name2 << " (age " << age2 <<").\n";
    else
        std::cout << name2 << " (age " << age2 <<") is older than " << name1 << " (age " << age1 <<").\n";
}
 
int main()
{
    const std::string name1{ getName(1) };
    const int age1 { getAge(name1) };
    
    const std::string name2{ getName(2) };
    const int age2 { getAge(name2) };
 
    printOlder(name1, age1, name2, age2);
 
    return 0;
}

问题 #5

In the solution to the above quiz, why can’t variable age1 in main be constexpr?

答案

A constexpr variable requires a constant expression initializer, and the call to getAge() isn’t allowed in a constant expression. Therefore, we can only make the variable const.