C++ Primer #02 Variables and Basic Types

2. Variables and Basic Types

2.1 Primitive Built-in Types

Exercise 2.3:

What output will the following code produce?

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unsigned u = 10, u2 = 42; 
std::cout << u2 - u << std::endl; 
std::cout << u - u2 << std::endl; 
int i = 10, i2 = 42; 
std::cout << i2 - i << std::endl; 
std::cout << i - i2 << std::endl; 
std::cout << i - u << std::endl; 
std::cout << u - i << std::endl;

---

Console:

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32
4294967264
32
-32
0
0

Exercise 2.5:

Determine the type of each of the following literals. Explain the differences among the literals in each of the four examples:

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(a) ’a’, L’a’, "a", L"a" 
​
(b) 10, 10u, 10L, 10uL, 012, 0xC 
​
(c) 3.14, 3.14f, 3.14L 
​
(d) 10, 10u, 10., 10e-2

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The answer is the following:

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(a): character literal, wide character literal, string literal, string wide character literal.
​
(b): decimal, unsigned decimal, long decimal, unsigned long decimal, octal, hexadecimal.
​
(c): double, float, long double.
​
(d): decimal, unsigned decimal, double, double.

👍 This is a good exercise which can explain everything in literals.

Exercise 2.6:

What, if any, are the differences between the following definitions

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int month = 9, day = 7; 
int month = 09, day = 07;

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Answer:

The first row is in decimal while the second row is in octal. Hence, int month = 09 is invalid which surpass 7(octal varies from 0-7).

Exercise 2.7:

What values do these literals represent? What type does each have?

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(a) "Who goes with F\145rgus?\012" 
(b) 3.14e1L
(c) 1024f 
(d) 3.14L

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(a) \145 means e , \012 means \n which changes line.

(b) 3.14e1L means long double. e1 means 10^1. Therefore, it will be 31.4.

(c) ⚠️ERROR. f stands for float. Hence, it should be 1024.0f.

(d) 3.14 Long Double.

2.2 Variables

Exercise 2.9:

Explain the following definitions. For those that are illegal, explain what’s wrong and how to correct it.

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(a) std::cin >> int input_value;
(b) int i = { 3.14 }; 
(c) double salary = wage = 9999.99; 
(d) int i = 3.14;

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(a) ⚠️ERROR. The variable should be initialized before feeding values. It should be:

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int input_value;
std::cin >> input_value;

(b) ⚠️Type double cannot be narrowed to int in initializer list. It should be:

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double i = {3.14};

(c) ⚠️ERROR. wage should be declared and initialized before salary. It should be:

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double wage = 700;
double salary = wage = 9000;
cout << wage << endl;
cout << salary << endl;

The wage and salary will have same value which are 9000.

(d) ✔️OK. i will be 3. Better solution should be:

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double i = 3.14;

Exercise 2.10:

What are the initial values, if any, of each of the following variables?

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std::string global_str; 
int global_int; 
int main() 
{
    int local_int; 
    std::string local_str;
}

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global_str is global variable, so the value is empty string.

global_int is global variable, so the value is zero.

local_int is a local variable which is uninitialized, so it has a undefined value.

local_str is also a local variable which is uninitialized, but it has a value that is defined by the class. So it is empty string.

‼️The RULE is Uninitialized objects of built-in type defined inside a function body have a undefined value. Objects of class type that we do not explicitly inititalize have a value that is defined by class.

Hence, the output of

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std::string global_str;
int global_int;
int main()
{
    int local_int;
    std::string local_str;
    cout << "--------------" << endl;
    cout << global_str << endl;
    cout << global_int << endl;
    cout << local_str << endl;
    cout << local_int << endl;
    cout << "--------------" << endl;
​
    return 0;
}

is :

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--------------
​
0
​
0
--------------

Exercise 2.11:

Explain whether each of the following is a declaration or a definition

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(a) extern int ix = 1024; 
(b) int iy; 
(c) extern int iz;

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(a) Definition.

(b) Definition.

(c) Declaration.

Exercise 2.12:

Which, if any, of the following names are invalid?

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(a) int double = 3.14;
(b) int _; 
(c) int catch-22; 
(d) int 1_or_2 = 1; 
(e) double Double = 3.14;

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(a) ❌

(b) ✔️. But it is not a good name.

(c) ❌

(d) ❌

(e) ✔️. But it is not a good variable name since it would be confused with the type ofdouble.

Exercise 2.13

What is the value of j in the following program?

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int i = 42; 
int main() 
{
    int i = 100; 
    int j = i;
}

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j is 100. Because the local i assigns its value to j.

Exercise 2.14:

Is the following program legal? If so, what values are printed?

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int i = 100, sum = 0; 
for (int i = 0; i != 10; ++i) 
    sum += i;
std::cout << i << " " << sum << std::endl;

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It should print out:

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100 45

The i inside the scope of main is a local variable. Hence, in the for loop, i != 100. While outside of scope, it is 100 .

2.3 Compound Types

Exercise 2.15:

Which of the following definitions, if any, are invalid? Why?

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(a) int ival = 1.01;
(b) int &rval1 = 1.01; 
(c) int &rval2 = ival; 
(d) int &rval3;

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(a) ✔️

(b) ❌ Does not reference any object.

(c) ✔️

(d) ❌ Does not initialized.

Exercise 2.16:

Which, if any, of the following assignments are invalid? If they are valid, explain what they do.

x
double d = 0, &r2 = d; 
int i = 0, &r1 = i; 
double d = 0, &r2 = d; 

x
(d) r1 = d;
(a) r2 = 3.14159;
(b) r2 = r1;
(c) i = r2;
(d) r1 = d;

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(a) ✔️. d also equals to 3.14

(b) ✔️. All are 0.

(c) ✔️. All are 0.

(d) ✔️. All are 0.

Exercise 2.17:

What does the following code print?

x
int i, &ri = i; i=5; ri= 10; 
std::cout << i << " " << ri << std::endl;

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Since ri is the reference of i, so i and ri are both equal to 10.

Exercise 2.18:

Write code to change the value of a pointer. Write code to change the value to which the pointer points.

---

x
#include <iostream>
​
using std::cout;
using std::cin;
using std::endl;
using std::cerr;
​
int main()
{
    int num = 10;
    int NUM = 20;
    int *pointer = &num;
​
    //change the value of a pointer
    pointer = &NUM;
    //change the value to which the pointer points.
    *pointer = 30;
​
    cout << "num: " << num << endl;
    cout << "NUM: " << NUM << endl;
    cout << "pointer: " << pointer << endl;
​
    return 0;
}

Console:

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num: 10
JBL: 30
pointer: 0x61fe10

Exercise 2.19:

Explain the key differences between pointers and references.

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1. A reference is another name of an already existing object. a pointer is an object in its own right.
2. Once initialized, a reference remains bound to its initial object. There is no way to rebind a reference to refer to a different object. a pointer can be assigned and copied.
3. a reference always get the object to which the reference was initially bound. a single pointer can point to several different objects over its lifetime.
4. a reference must be initialized. a pointer need not be initialized at the time it is defined.

Exercise 2.20:

What does the following program do?

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int i = 42; 
int *p1 = &i; *p1 = *p1 **p1;

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The program changes the valuei pointed by p1 to 42*42 by dereference.

Exercise 2.21:

Explain each of the following definitions. Indicate whether any are illegal and, if so, why

x
int i = 0; 

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(a) double* dp = &i; 
(b) int *ip = i; 
(c) int *p = &i;

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(a) ❌Different type. More technical definition is: cannot initialize a variable of type double * with an rvalue of type int *

(b) ❌ Invalid syntax. More technical definition is: cannot initialize a variable of type int * with an lvalue of type int

(c) ✔️ OK.

Exercise 2.22:

Assuming p is a pointer to int, explain the following code:

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if (p) // ... 
if (*p) // ...

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(1) First line: whether p is nullptr? So there may be 2 solutions: true - p is not nullptr, false - p is nullptr.

(2) Second line: whether *p is zero? So there may be 2 solutions: true - *p is not zero, false - *p is zero.

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#include <iostream>
​
using std::cout;
using std::cin;
using std::endl;
using std::cerr;
​
int main()
{
​
    int i=10;
    int *j = nullptr; //equivalent to int *j=0;
    int *k = &i;
​
    if(i)
    {
        cout << "i out" << endl;
    }
    if(j)
    {
        cout << "j out" << endl; //nullptr=false
    }
    if(k)
    {
        cout << "k out" << endl;
    }
​
    return 0;
}
​

Console:

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i out
k out
​
Process returned 0 (0x0)   execution time : 1.455 s
Press any key to continue.

Hence, nullptr is false.

Exercise 2.24:

Why is the initialization of p legal but that of lp illegal?

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int i = 42; 
void *p = &i; 
long *lp = &i;

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Because void * is special which can the address of any object. While long * cannot.

Exercise 2.25:

Determine the types and values of each of the following variables.

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(a) int* ip, i, &r = i; 
(b) int i, *ip = 0; 
(c) int* ip, ip2;

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(a) ip is a pointer. i is an int with undefined value. r is a reference of i.

(b) i is an int with undefined value. ip is a pointer, especially nullptr.

(c) ip is a pointer. ip2 is an int.

2.4 const Qualifier

Exercise 2.26:

Which of the following are legal? For those that are illegal, explain why.

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(a) const int buf;
(b) int cnt = 0;
(c) const int sz = cnt;
(d) ++cnt; ++sz;