CS100 Lecture 12

References, std::vector

Contents

• References
• std::vector

References

Declare a reference

A reference defines an alternative name for an object ("refers to" that object).

Similar to pointers, the type of a reference is ReferredType &, which consists of two things:

• ReferredType is the type of the object that it refers to, and
• & is the symbol indicating that it is a reference.

Example:

int ival = 42;
int &ri = ival; // `ri` refers to `ival`.
                // In other words, `ri` is an alternative name for `ival`.
std::cout << ri << '\n'; // prints the value of `ival`, which is `42`.
++ri;           // Same effect as `++ival;`.

Declare a reference

int ival = 42;
int x = ival;              // `x` is another variable.
++x;                       // This has nothing to do with `ival`.
std::cout << ival << '\n'; // 42
int &ri = ival;            // `ri` is a reference that refers to `ival`.
++ri;                      // This modification is performed on `ival`.
std::cout << ival << '\n'; // 43

Ordinarily, when we initialize a variable, the value of the initializer is copied into the object we are creating.

When we define a reference, instead of copying the initializer's value, we bind the reference to its initializer.

A reference is an alias

When we define a reference, instead of copying the initializer's value, we bind the reference to its initializer.

int ival = 42;
int &ri = ival;
++ri;           // Same as `++ival;`.
ri = 50;        // Same as `ival = 50;`.
int a = ri + 1; // Same as `int a = ival + 1;`.

After a reference has been defined, all operations on that reference are actually operations on the object to which the reference is bound.

ri = a;

What is the meaning of this?

A reference is an alias

int ival = 42;
int &ri = ival;
++ri;           // Same as `++ival;`.
ri = 50;        // Same as `ival = 50;`.
int a = ri + 1; // Same as `int a = ival + 1;`.

When we define a reference, instead of copying the initializer's value, we bind the reference to its initializer.

After a reference has been defined, all operations on that reference are actually operations on the object to which the reference is bound.

ri = a;

• This is the same as ival = a;. It is not rebinding ri to refer to a.

A reference must be initialized

ri = a;

• This is the same as ival = a;. It is not rebinding ri to refer to a.

Once initialized, a reference remains bound to its initial object. There is no way to rebind a reference to refer to a different object.

Therefore, references must be initialized.

References must be bound to existing objects ("lvalues")

It is not allowed to bind a reference to temporary objects or literals :

int &r1 = 42;    // Error: binding a reference to a literal
int &r2 = 2 + 3; // Error: binding a reference to a temporary object
int a = 10, b = 15;
int &r3 = a + b; // Error: binding a reference to a temporary object

In fact, the references we learn today are "lvalue references", which must be bound to lvalues. We will talk about value categories in later lectures.

References are not objects

A reference is an alias. It is only an alternative name of another object, but the reference itself is not an object.

Therefore, there are no "references to references".

int ival = 42;
int &ri = ival; // binding `ri` to `ival`.
int & &rr = ri; // Error! No such thing!

What is the meaning of this code? Does it compile?

int &ri2 = ri;

References are not objects

A reference is an alias. It is only an alternative name of another object, but the reference itself is not an object.

Therefore, there are no "references to references".

int ival = 42;
int &ri = ival; // binding `ri` to `ival`.
int & &rr = ri; // Error! No such thing!

What is the meaning of this code? Does it compile?

int &ri2 = ri; // Same as `int &ri2 = ival;`.

• ri2 is a reference that is bound to ival.
• Any use of a reference is actually using the object that it is bound to!

References are not objects

A reference is an alias. It is only an alternative name of another object, but the reference itself is not an object.

Pointers must also point to objects. Therefore, there are no "pointers to references".

int ival = 42;
int &ri = ival; // binding `ri` to `ival`.
int &*pr = &ri; // Error! No such thing!

What is the meaning of this code? Does it compile?

int *pi = &ri;

References are not objects

A reference is an alias. It is only an alternative name of another object, but the reference itself is not an object.

Pointers must also point to objects. Therefore, there are no "pointers to references".

int ival = 42;
int &ri = ival; // binding `ri` to `ival`.
int &*pr = ri; // Error! No such thing!

What is the meaning of this code? Does it compile?

int *pi = &ri; // Same as `int *pi = &ival;`.

Reference declaration

Similar to pointers, the ampersand & only applies to one identifier.

int ival = 42, &ri = ival, *pi = &ival;
// `ri` is a reference of type `int &`, which is bound to `ival`.
// `pi` is a pointer of type `int *`, which points to `ival`.

Placing the ampersand near the referred type does not make a difference:

int& x = ival, y = ival, z = ival;
// Only `x` is a reference. `y` and `z` are of type `int`.

* and &

Both symbols have many identities!

• In a declaration like Type *x = expr, * is a part of the pointer type Type *.
• In a declaration like Type &r = expr, & is a part of the reference type Type &.
• In an expression like *opnd where there is only one operand, * is the dereference operator.
• In an expression like &opnd where there is only one operand, & is the address-of operator.
• In an expression like a * b where there are two operands, * is the multiplication operator.
• In an expression like a & b where there are two operands, & is the bitwise-and operator.

Example: Use references in range-for

Recall the range-based for loops (range-for):

std::string str;
std::cin >> str;
int lower_cnt = 0;
for (char c : str)
  if (std::islower(c))
    ++lower_cnt;
std::cout << "There are " << lower_cnt << " lowercase letters in total.\n";

The range-for loop in the code above traverses the string, and declares and initializes the variable c in each iteration as if

for (std::size_t i = 0; i != str.size(); ++i) {
  char c = str[i]; // Look at this!
  if (std::islower(c))
    ++lower_cnt;
}

Example: Use references in range-for

for (char c : str)
  // ...

The range-for loop in the code above traverses the string, and declares and initializes the variable c in each iteration as if

for (std::size_t i = 0; i != str.size(); ++i) {
  char c = str[i];
  // ...
}

Here c is a copy of str[i]. Therefore, modification on c does not affect the contents in str.

Example: Use references in range-for

What if we want to change all lowercase letters to their uppercase forms?

for (char c : str)
  c = std::toupper(c); // This has no effect.

We need to declare c as a reference.

for (char &c : str)
  c = std::toupper(c);

This is the same as

for (std::size_t i = 0; i != str.size(); ++i) {
  char &c = str[i];
  c = std::toupper(c); // Same as `str[i] = std::toupper(str[i]);`.
}

Example: Pass by reference-to-const

Write a function that accepts a string and returns the number of lowercase letters in it:

int count_lowercase(std::string str) {
  int cnt = 0;
  for (char c : str)
    if (std::islower(c))
      ++cnt;
  return cnt;
}

To call this function:

int result = count_lowercase(my_string);

Example: Pass by reference-to-const

int count_lowercase(std::string str) {
  int cnt = 0;
  for (char c : str)
    if (std::islower(c))
      ++cnt;
  return cnt;
}

int result = count_lowercase(my_string);

When passing my_string to count_lowercase, the parameter str is initialized as if

std::string str = my_string;

The contents of the entire string my_string are copied!

Example: Pass by reference-to-const

int result = count_lowercase(my_string);

When passing my_string to count_lowercase, the parameter str is initialized as if

std::string str = my_string;

The contents of the entire string my_string are copied! Is this copy necessary?

Example: Pass by reference-to-const

int result = count_lowercase(my_string);

When passing my_string to count_lowercase, the parameter str is initialized as if

std::string str = my_string;

The contents of the entire string my_string are copied! This copy is unnecessary, because count_lowercase is a read-only operation on str.

How can we avoid this copy?

Example: Pass by reference-to-const

int count_lowercase(std::string &str) { // `str` is a reference.
  int cnt = 0;
  for (char c : str)
    if (std::islower(c))
      ++cnt;
  return cnt;
}

int result = count_lowercase(my_string);

When passing my_string to count_lowercase, the parameter str is initialized as if

std::string &str = my_string;

Which is just a reference initialization. No copy is performed.

Example: Pass by reference-to-const

int count_lowercase(std::string &str) { // `str` is a reference.
  int cnt = 0;
  for (char c : str)
    if (std::islower(c))
      ++cnt;
  return cnt;
}

However, this has a problem:

std::string s1 = something(), s2 = some_other_thing();
int result = count_lowercase(s1 + s2); // Error: binding reference to
                                       // a temporary object.

a + b is a temporary object, which str cannot be bound to.

Example: Pass by reference-to-const

References must be bound to existing objects, not literals or temporaries.

There is an exception to this rule: References-to-const can be bound to anything.

const int &rci = 42; // OK.
const std::string &rcs = a + b; // OK.

rcs is bound to the temporary object returned by a + b as if

std::string tmp = a + b;
const std::string &rcs = tmp;

We will talk more about references-to-const in recitations.

Example: Pass by reference-to-const

The answer:

int count_lowercase(const std::string &str) { // `str` is a reference-to-`const`.
  int cnt = 0;
  for (char c : str)
    if (std::islower(c))
      ++cnt;
  return cnt;
}

std::string a = something(), b = some_other_thing();
int res1 = count_lowercase(a);       // OK.
int res2 = count_lowercase(a + b);   // OK.
int res3 = count_lowercase("hello"); // OK.

Benefits of passing by reference-to-const

Apart from the fact that it avoids copy, declaring the parameter as a reference-to-const also prevents some potential mistakes:

int some_kind_of_counting(const std::string &str, char value) {
  int cnt = 0;
  for (std::size_t i = 0; i != str.size(); ++i) {
    if (str[i] = value) // Ooops! It should be `==`.
      ++cnt;
    else {
      // do something ...
      // ...
    }
  }
  return cnt;
}

str[i] = value will trigger a compile-error, because str is a reference-to-const.

Benefits of passing by reference-to-const

1. Avoids copy.
2. Accepts temporaries and literals (rvalues).
3. The const qualification prevents accidental modifications to it.

[Best practice] Pass by reference-to-const if copy is not necessary and the parameter should not be modified.

References vs pointers

Both a reference and a pointer can be used to refer to an object, but references are more convenient - no need to write the annoying * and &.

Note: nullptr is the null pointer value in C++. Do not use NULL.

std::vector

Defined in the standard library file <vector>.

A "dynamic array".

Class template

std::vector is a class template.

Class templates are not themselves classes. Instead, they can be thought of as instructions to the compiler for generating classes.

• The process that the compiler uses to create classes from the templates is called instantiation.

For std::vector, what kind of class is generated depends on the type of elements we want to store, often called value type. We supply this information inside a pair of angle brackets following the template's name:

std::vector<int> v; // `v` is of type `std::vector<int>`

Create a std::vector

std::vector is not a type itself. It must be combined with some <T> to form a type.

std::vector v;               // Error: missing template argument.
std::vector<int> vi;         // An empty vector of `int`s.
std::vector<std::string> vs; // An empty vector of strings.
std::vector<double> vd;      // An empty vector of `double`s.
std::vector<std::vector<int>> vvi; // An empty vector of vector of `int`s.
                                   // "2-d" vector.

What are the types of vi, vs and vvi?

Create a std::vector

std::vector is not a type itself. It must be combined with some <T> to form a type.

std::vector v;               // Error: missing template argument.
std::vector<int> vi;         // An empty vector of `int`s.
std::vector<std::string> vs; // An empty vector of strings.
std::vector<double> vd;      // An empty vector of `double`s.
std::vector<std::vector<int>> vvi; // An empty vector of vector of `int`s.
                                   // "2-d" vector.

What are the types of vi, vs and vvi?

• std::vector<int>, std::vector<std::string>, std::vector<std::vector<int>>.

Create a std::vector

There are several common ways of creating a std::vector:

std::vector<int> v{2, 3, 5, 7};     // A vector of `int`s,
                                    // whose elements are {2, 3, 5, 7}.
std::vector<int> v2 = {2, 3, 5, 7}; // Equivalent to ↑

std::vector<std::string> vs{"hello", "world"}; // A vector of strings,
                                    // whose elements are {"hello", "world"}.
std::vector<std::string> vs2 = {"hello", "world"}; // Equivalent to ↑

std::vector<int> v3(10);     // A vector of ten `int`s, all initialized to 0.
std::vector<int> v4(10, 42); // A vector of ten `int`s, all initialized to 42.

Note that all the elements in v3 are initialized to 0.

• We hate uninitialized values, so does the standard library.

Create a std::vector

Create a std::vector as a copy of another one:

std::vector<int> v{2, 3, 5, 7};
std::vector<int> v2 = v; // `v2`` is a copy of `v`
std::vector<int> v3(v);  // Equivalent
std::vector<int> v4{v};  // Equivalent

No need to write a loop!

Copy assignment is also enabled:

std::vector<int> v1 = something(), v2 = something_else();
v1 = v2;

• Element-wise copy is performed automatically.
• Memory is allocated automatically. The memory used to store the old data of v1 is deallocated automatically.

C++17 CTAD

"Class Template Argument Deduction": As long as enough information is supplied in the initializer, the value type can be deduced automatically by the compiler.

std::vector v1{2, 3, 5, 7}; // vector<int>
std::vector v2{3.14, 6.28}; // vector<double>
std::vector v3(10, 42);     // vector<int>, deduced from 42 (int)
std::vector v4(10);         // Error: cannot deduce template argument type

Size of a std::vector

v.size() and v.empty(): same as those on std::string.

std::vector v{2, 3, 5, 7};
std::cout << v.size() << '\n';
if (v.empty()) {
  // ...
}

v.clear(): Remove all the elements.

Append an element to the end of a std::vector

v.push_back(x)

int n;
std::cin >> n;
std::vector<int> v;
for (int i = 0; i != n; ++i) {
  int x;
  std::cin >> x;
  v.push_back(x);
}
std::cout << v.size() << '\n'; // n

Remove the last element of a std::vector

v.pop_back()

Exercise: Given v of type std::vector<int>, remove all the consecutive even numbers in the end.

Remove the last element of a std::vector

v.pop_back()

Exercise: Given v of type std::vector<int>, remove all the consecutive even numbers in the end.

while (!v.empty() && v.back() % 2 == 0)
  v.pop_back();

v.back(): returns the reference to the last element.

• How is it different from "returning the value of the last element"?

v.back() and v.front()

Return the references to the last and the first elements, respectively.

It is a reference, through which we can modify the corresponding element.

v.front() = 42;
++v.back();

For v.back(), v.front() and v.pop_back(), the behavior is undefined if v is empty. They do not perform any bounds checking.

Range-based for loops

A std::vector can also be traversed using a range-based for loop.

std::vector<int> vi = some_values();
for (int x : vi)
  std::cout << x << std::endl;
std::vector<std::string> vs = some_strings();
for (const std::string &s : vs) // use reference-to-const to avoid copy
  std::cout << s << std::endl;

Exercise: Use range-based for loops to count the number of uppercase letters in a std::vector<std::string>.

Range-based for loops

Exercise: Use range-based for loops to count the number of uppercase letters in a std::vector<std::string>.

int cnt = 0;
for (const std::string &s : vs) { // Use reference-to-const to avoid copy
  for (char c : s) {
    if (std::isupper(c))
      ++cnt;
  }
}

Access through subscripts

v[i] returns the reference to the element indexed i.

• i , where v.size().
• Subscript out of range is undefined behavior. v[i] performs no bounds checking.
  • In pursuit of efficiency, most operations on standard library containers do not perform bounds checking.
• A kind of "subscript" that has bounds checking: v.at(i).
  • If i is out of range, a std::out_of_range exception is thrown.

Feel the style of STL

Basic and low-level operations are performed automatically:

• Default initialization of std::string and std::vector results in an empty string / container, not indeterminate values.
• Copy of std::string and std::vector is done automatically, which performs member-wise copy.
• Memory management is done automatically.

Interfaces are consistent:

• std::string also has member functions like .push_back(x), .pop_back(), .at(i), .size(), .clear(), etc. which do the same things as on std::vector.
• Both can be traversed by range-for.

Summary

References

• A reference is an alias.
• A reference is bound to an object during initialization. After that, any use of that reference is actually using the object it is bound to.
• A reference can only be bound to existing objects (lvalues). A pointer can only point to existing objects.
  • But a reference-to-const can be bound to anything.
• Pass arguments by reference-to-const: avoids copy, accepts both lvalues and rvalues, and prevents accidental modification on what should not be modified.

Summary

std::vector

• std::vector is not a type. It must be combined with some <T> to form a type.
• Many ways of creation.
• Copy of a std::vector performs member-wise copy.
• v.size, v.empty, v.push_back, v.pop_back, v.clear, v[i], v.at(i).
• Use range-for to traverse a std::vector.

Exercises

Write the exercises on page 26, 38, 40 and 43 on your own.

Notes

String literals ("hello") are an exception to this. Integer literals, floating-point literals, character literals, boolean literals and enum items are rvalues, but string literals are lvalues. They do live somewhere in the memory.

In fact, the range-for uses iterators, not subscripts.