C++ Review

  • Below are the list of topics covered in this review.
  • This document is not by any stretch a comprehensive reference of cpp,
  • it just covers some of the topics that we will be using in our course only.
  • All the code here are can be imported into MSVC as a single solution from this repository.
  1. Variables and References
  2. Classes
  3. Move semantics
  4. Return values
  5. Pointers
  6. Templates
  7. STL

Variables and references

  • When we define (declare) a variable the system reserves space in memory to store the
    value associated with that variable.
  • This is the reason why we need to specify the type of the variable since the required space depends on it.
  • For example (typically), an int and float need 4 bytes whereas long long and double need 8 bytes.
  • We can determine the memory address where the variable is located using the & operator.
  • Note that the & operator can have different meaning depending on context.

Example.

#include <iostream>
int main(){
//a location in memory is reserved and labeled x
int x=2;
//y is just another name for the same location. no reservation is done.
int& y=x;
//a location is reserved for z and the value of x is copied
int z=x;
z=17;
y=13;
//print the value of the variables and their respective addresses
std::cout<<"x= "<<x<<" and &x="<<&x<<std::endl;
std::cout<<"y= "<<y<<" and &y="<<&y<<std::endl;
std::cout<<"z= "<<z<<" and &z="<<&z<<std::endl;

}
  • Note that int& y=x; declares y as a reference to x
  • whereas &x gives the memory address of x.
  • The different declarations used above carry to the parameters in function calls.
  • For example,
#include <iostream>
void byValue(int n){
    n=17;
}
void byRef(int& n){
    n=12;
}
int main(){
  int x=2;
  byValue(x);
  std::cout<<x<<std::endl;
  byRef(x);
  std::cout<<x<<std::endl;

}
  • In the call to the function byValue(x) it is as if we declare int n=x; and therefore n is a copy of x.
  • By contrast, byRef(x) is is as if we declare int& n=x; so no copy is made and n is a reference to x.
  • Usually we call by reference when either we want to change the input or when the input is large and copying becomes expensive.
  • We can use the best of both by using a const reference
int byCRef(const int& n){
    n=37;//error cannot modify n
    return 2*n;
}
  • Also, const allows us to pass literals and temporaries.
int byT(int n){
    return 7*n;
}
int byCRef(const int & n){
    n=n+1;//error n is const
    return 2*n;
}
int byRef(int& n){
    n=n+1;//changes the value of parameter
    return 2*n;
}
int main(){
    byRef(2);//error cannot bind a non-const lvalue to rvalue
    byCRef(2);//OK
    byRef(byT(2));//error since the return value of byT is a temp
    byCRef(byT(2));//OK
    int& r=byT(2);//error cannot bind 
    int&& res=byT(2);//ok
}

You can run the above code here.

  • One important property of references is that they cannot be reassigned.
#include <iostream>
int main(){
int x=19,y=23;

//int &xr;//error a reference must be initialized
int& xr=x;
int& yr=y;
/* this does NOT reassign xr to reference y
* it merely assigns the value of y to xr and thus to x
*/
xr=yr;
std::cout<<"value of x="<<x<<"\n";
xr=45;
std::cout<<"value of x="<<x<<"\n";
std::cout<<"value of y="<<y<<"\n";
}

You can try it here

  • We can prevent modification to a variable we reference to.

  • For example in the above if we declare const int& xr=x; it means that x cannot be modified using xr (it can still be modified).

  • In this case the line xr=yr; will generate an error.

  • Note that in some text they write int const& xr=x;

  • This is equivalent but the first syntax conveys the fact that the int is unmodifiable not the reference, since by definition a reference is unmodifiable.

  • As we will see later a class containing a member of type reference cannot be assigned (assignment operator is deleted)

Classes

  • In C++ new types are created using classes.
  • Once a class is defined new objects can be instantiated from such a class.
  • Minimal syntax of a (useless) class
class Test{};
int main(){
    Test t;
}
  • A class can have member variables and member functions.
class Test{
    int _x;
    public:
    int& x(){
        return _x;
    }
    
};
int main(){
    Test t;// at this point _x is undefined
    t.x()=17;
}
  • By default all members of a class are private and hence inaccessible from outside the scope of the object.
  • To make a member accessible we use the keyword public.
  • Note that the member function x() returns a reference to _x and this allows us to change the value of _x.
  • We can use pointers as usual where the code below is equivalent to the one above but using the arrow instead of the dot operator.
int main(){
    Test * p=new Test();
    p->x()=17;
}

In fact the private and public qualifiers can be used for any and all members. For example

class Test {
    private: int _x;//_x is private
    public: int _y;// _y is public
     int _z;// _z is public. The keyword carries over until it changes
     private: void f(){}
     public: int& x(){return _x;}
}
  • But usually all public members are grouped together using a single keyword and the same for private members;
int main(){
    Test t;
    t._x;//error _x is inaccessible
    t._y=t._z;//OK both are public
    t.x();//OK
    t.f();//Error f is inaccessible
}

Constructors and destructors

  • For builtin types like int and double a variable is "created" (memory is reserved) when the variable is declared.
  • Once the variable is out of scope is it "destroyed" (memory is released).
  • The same thing is done for objects instantiated from classes.
  • This is done by using constructor and destructor.
  • When we don't supply our own versions a default version is used by the compiler which basically calls the constructors and destructors of the member variables.
class Test {
    public:
    int _x;
    double _y;
}
int main(){
    Test t;
    std::cout<<t._x<<std::endl;
    std::cout<<t._y<<std::endl;
}

No constructor is supplied so the compiler uses a default that creates variables _x and _y.

A constructor builds an object bottom up.

  1. the constructor of the base class (if any) is called
  2. members instructors are called
  3. Finally the constructor body is executed.

For the destructor the opposite happens.
For example

struct Item {
    Item(){std::cout<<"Item ctor\n";}
    Item(int i){std::cout<<"Item ctor with input\n";}
};
struct Test {
    Item _i;
    int x;
};
void noinit(){
    int x=12,y=77,z=99;
    Test t;
    std::cout<<t.x<<std::endl;
}
void init(){
    int x=12,y=77,z=99;
    Test t {};//initialize to zero
    std::cout<<t.x<<std::endl;
}
int main(){
    noinit();
    init();
}

Run to get the output

item ctor
723520304
item ctor
0
  • As we can see from the above example built-in types are not initiaized
  • sometimes they are zero sometimes they are not, it depends on the compiler.
  • For class types the default constructor is called. We can control the constructor and the initialization of members as follows
struct Test {
    Test(int x,int i):_x(x),_i(i){}
}
  • As mentioned before, classes containing references or constants have their assignment operator automatically deleted
#include <iostream>
struct TestRef {
    int& _x;
    TestRef(int& x) :_x(x) {}
};
struct TestConst {
    const int _x;
    TestConst(int x) :_x(x) {}
};
int main()
{
    int x = 12;
    TestRef t(x);
    TestRef p(x);
    TestConst c(x);
    TestConst d(x);
    p = t;//error assignment operator deleted
    c = d;//error assignment operator deleted
}

You can see the errors here

Rvalue reference and move semantics

  • C++11 introduced rvalue references. Variable res below extends the lifetime of the temporary object created by the RT() function.
  • Consider when the destructor is called below
struct Test {
  int _x;
  Test(int x=0):_x(x){}
  ~Test(){
    std::cout<<"dtor "<<_x<<std::endl;
  }
};
Test RT(int val){
   return Test(val);
}
int main() {
Test&& res=RT(8);
std::cout<<"creating 7\n";
RT(7);
std::cout<<" done\n";
}

-You can run the above code here

  • Note that when an rvalue reference is used, it is used as a lvalue reference.
  • This is called move semantics is not passed through .
  • For the example the following recursive function gives an error
void doit(std::string&& s){
  if(s!="hello")
    doit(s);

}

this is a fix

void doit(std::string&& s){
  if(s!="hello"){
    s="hello"; //this line so we don't go into infinite recursion
    doit(std::move(s));
  }
    
}

Move semantics allows us to transfer ownership of resources.

int main(){
std::string s{"hello there"};
std::string ts=std::move(s);
std::cout<<"the string s is "<<s<<"\n";
std::cout<<"the string ts is "<<ts<<"\n";
std::vector<int> v{1,2,3,4};
std::cout<<"values of v before move are ";
for(auto& x:v)std::cout<<x<<",";
std::vector<int> tv=std::move(v);
std::cout<<"\nafter move they are ";
for(auto& x:v)std::cout<<x<<",";
std::cout<<"\nand the values of tv are ";
for(auto& x:tv)std::cout<<x<<",";
std::cout<<std::endl;

}

you can try it here. We illustrate further with our own, very simple, container.

struct Container {
    int* p = nullptr;
    int _size;
    Container(int size) :_size(size), p(new int[size] {}) {}
    Container(const Container& rhs) {//copy constructor
        if (&rhs != this) {
            _size = rhs._size;
            p = new int[_size];
            for (int i = 0; i < _size; ++i)
                p[i] = rhs.p[i];
        }
    }
    //usually there is also an assignment operator
    Container(Container&& rhs) {
        p = rhs.p;
        rhs.p = nullptr;
        rhs._size = 0;
    }
    Container& operator=(Container&& rhs) {
        if (p)delete[] p;
        p = rhs.p;
        _size = rhs._size;
        rhs._size = 0;
        rhs.p = nullptr;
        return *this;
    }
    void print() {
        for (int i = 0; i < _size; ++i)
            std::cout << p[i] << ",";
        std::cout << "\n";
    }
    ~Container() {
        if (p)delete[] p;
    }
};

int main()
{    
    Container c(10);
    std::cout<<"Content of c \n";
    c.print();
    Container d(std::move(c));
    std::cout<<"Content of c \n";
    c.print();

    Container e(3);
    d = std::move(e);
    std::cout<<"Content of d \n";
    d.print();
}

You can try it here

Return values

  • unless the compiler performs return value optimization (rvo) the following occurs
  • (in g++ or clang++ specify -fno-elide-constructors to skip optimization)
struct Test {
        Test(){ std::cout<<"ctor\n";}
        Test(const Test& rhs){
          std::cout<<"copy ctor\n";
        }
        ~Test(){ std::cout<<"dtor\n";}
};
Test retTest(){return Test();}
int main(){
  Test t=retTest();
  std::cout<<"temporary returned by refTest is destroyed\n";
  std::cout<<"end of main t will be destroyed\n";
}

what happens is the following

  1. inside function retTest() an object of type Test is created on the stack
  2. a tmp object of type Test is copy constructed from that object
  3. the object on the stack is dtored
  4. t in main is copy ctored from the tmp
  5. tmp is destroyed
  6. when main exists t is destroyed

You can test the code below here. Note the -fno-elide-constructors option in the bottom right of the screen.

$g++-10 -fno-elide-constructors -std=c++11 rvopt.cpp
$./a.out
ctor
copy ctor
dtor
copy ctor
dtor
temporary returned by refTest is destroyed
end of main t will be destroyed
dtor

If we remove the -fno-elide-constructors you get this output. Try it here

$g++-10 -std=c++20 rvopt.cpp
$./a.out
ctor
temporary returned by refTest is destoryed
end of main t will be destroyed
dtor

Note: since c++17 this is no longer possible. Try it here

Pointers

A pointer variable is a variable that holds an address. We say variable p points to variable x if p holds the address of x: int *p=&x;.

int main(){
int x=17,y=45;
int* p=&x;
std::cout<<p<<std::endl;//prints the value of p, i.e. the address of x
std::cout<<*p<<std::endl;//prints the value store at the location p, i.e. x
*p=23;//change the value of x
p=&y;//p now stores the address of y
}

Pointers usually are used when we need to dynamically allocate memory.

int main(){
    int *p=new int;//reserve space for int. Value undefined
    int *q=new int(8);//reserve space for int and store 8
    *p=55;//store value 55 at address p
    delete p;//release the reserved memory;
}

The new expression can be used with any object. In particular, we can use it to create an array dynamically

int main(){
    const int n=8;
    int* p=new int[n];//Note the difference from int(n)
    /* fill the array with values */
    /* p IS a variable so we make
     * copy before changing it
     */
    for(int i=0,*q=p;i<n;++i,++q)
         *q=i;
    for(int i=0;i<n;++i)
      std::cout<<p[i]<<",";
    std::cout<<"\n";

}

As we mentioned before we can use a pointer to any type.

class Test {
int _x,_y;
public:
 Test(int x,int y):_x(x),_y(y){}
 int& getX(){return _x;}
 int& getY(){return _y;}
};
int main(){
    Test* t=new Test(13,18);
    t->getX()=3;
    t->getY()=7;
    std::cout<<++t->getX()<<"\n";
    std::cout<<++t->getY()<<"\n";
}

You can try the above code here

new, malloc, operator new

  • A new expression is used both for dynamically allocating memory(on the heap) and calling the constructor of an object.
  • The function operator new allocates memory only.
  • In that sense it is similar to malloc in C.
  • Unless you are designing your own container you almost never need to use operator new.
  • Usually it is used to place the constructed object at a preallocated place.

Example

#include <new>
struct Test {
 int _x,_y;
 Test(int x,int y):_x(x),_y(y){std::cout<<"ctor\n";}
 ~Test(){std::cout<<"dtor\n";}
};
int main(){
void* p=operator new(sizeof(Test));
std::cout<<"finished allocating memory\n";
Test* t=new (p) Test{1,2};
delete t;
}

You can run the code here.

  • We can override the implementation of operator new and operator delete.
  • As can be seen below new and delete are the C++ "versions" of C malloc and free.
#include <iostream>
class Test {
int _x,_y;
public:
 Test(int x,int y):_x(x),_y(y){std::cout<<"ctor\n";}
 int& getX(){return _x;}
 int& getY(){return _y;}
 ~Test(){std::cout<<"dtor\n";}
};


void * operator new(std::size_t size){
   std::cout<<"allocating size ="<<size<<" \n";
   void *p=malloc(size);
   return p;

}
void operator delete(void *p) noexcept {
    std::cout<<"freeing memory\n";
    free(p);
}
int main(){
    Test *t=new (Test){1,2};
    delete t;
    Test *p=new Test{3,4};
    p->~Test();
    operator delete(p);
}

You can try the above here

Smart pointers

  • pointers provide flexibility but they can cause a variety of problems.
  • One of the problems is shared ownership of a resource.
  • In order not to have memory leaks, we need to free the allocated memory when done.
  • This particular problem occurs when two or more pointer variables have the address of the same dynamically allocated resource.
  • In such a situation we either try to free the memory multiple times, access a resource that no longer exists
  • or forget to release the memory which causes memory leaks.

We illustrate with the following simple example.

using Long = unsigned long long;

void* memory_block(size_t size) {
	return operator new(size);
}
void release_memory(void* p) {
	operator delete(p);
}

struct Leaker {
	int* _values=nullptr;
	int _size;
	Leaker(Long size) :_size(size) {
		_values = (int*)memory_block(_size * sizeof(int));
	}
	int* values() {
		return _values;
	}
    ~Leaker(){//Do we free memory here ?
    }
};

int main() {
	const unsigned long long n = 1 << 20;
	for (int i = 0; i < 50; ++i) {
		Leaker x(n);
		int* p = x.values();
    }
    /* keep the program running long enough */
	int y;
	std::cout << "type anything\n";
	std::cin >> y;
}
  • the class Leaker allocates 1MB of memory.
  • It has a choice, either it does not free the allocated memory, as we did in the example above
  • or frees it in the destructor with the danger that p will also free it.
  • Using the performance profiler in VS (Debug->Performance profiler) we see that the above code uses about 200MB
  • which makes sense since each iteration allocates 4MB.

Figure 1

The other choice is to modify the destructor as follows

~Leaker(){
    if(_values!=nullptr)delete _values;
}

This works provided the user (i.e. the code calling int p*=x.value()) does not execute delete p; which crashes the program. You can try that scenario here

  • To avoid problems like these we use either std::unique_ptr<T> or std::shared_ptr<T>.
  • The first enforces exclusive
    ownership whereas
  • the second allows shared ownership.
  • We start with an example of the second.
    using Long = unsigned long long;

void* memory_block(size_t size) {
	return operator new(size);
}
void release_memory(void* p) {
	operator delete(p);
}
struct Shared_Owner {
	std::shared_ptr<int> _values;
	Long _size;
	Shared_Owner(Long size) :_size(size) {
		/* in three steps for clarity */
		void* raw = memory_block(_size * sizeof(int));
		std::shared_ptr<int> q((int*)raw);
		_values = q;
		/*q will be destroyed here. It is ok
		* it holds no resource since it was
		* transfered to p
		*/
	}
	std::shared_ptr<int> values() {
		return _values;
	}
    long count(){
        return _values.use_count();
    }
};

int main(){
    const unsigned long long n = 1 << 20;
    Shared_Owner x(n);

	{
		std::shared_ptr<int> p = x.values();
        std::cout<<"after p is created="<<x.count()<<"\n";
        std::shared_ptr<int> q=x.values();
        std::cout<<"after q is created="<<x.count()<<"\n";
    }
    std::cout<<"count outside block= "<<x.count()<<"\n";
}
  • there is no delete of a std::shared_ptr - - it is automatically destroyed (i.e. dtor is called) when it goes out of scope and
  • it frees the resource it is pointing to only if it is the last shared_ptr copy.
  • it is used exactly like pointers.
  • In the example above there are three std::shared_ptr,
  • all pointing to the resource allocated by the Shared_Owner object x.
  • Inside the block two std::shared_ptr objects are created p and q,
  • both pointing to the memory block allocated by x. - When they go out of scope, and therefore they are destroyed,
  • the allocated memory is not freed, the number of copies is just decremented.
  • when x goes out of scope, it is the last shared_ptr pointing to the resource
  • therefore it frees it.
    -You can try the above example here here.
  • The std::unique_ptr<T> is similar except it enforces exclusive ownership.
  • Below is a similar example illustrating the memory automatically released by the std::unique_ptr when it is destroyed.
  • Note the two changes due to noncopiable nature of std::unique_ptr.
  • First, in function mod the std::unique_ptr must be passed and returned by reference
  • we cannot make a copy.
  • in the statement std::unique_ptr<int> t = std::move(mod(p->res)); the resource held by p is transferred to t.
using Long = unsigned long long;

void* memory_block(size_t size) {
	return operator new(size);
}
void release_memory(void* p) {
	operator delete(p);
}

struct Unique_Owner {
	std::unique_ptr<int> _values;
	Long _size;
	Unique_Owner(Long size) :_size(size) {
		/* in three steps for clarity */
		void* raw = memory_block(_size * sizeof(int));
		std::unique_ptr<int> q ((int *)raw);
		_values = std::move(q);
		/*q will be destroyed here. It is ok 
		* it holds no resource since it was
		* transfered to p
		*/
	}
	std::unique_ptr<int> values() {
		return std::move(_values);
	}
};

int main(){
    const unsigned long long n = 1 << 20;
	for (int i = 0; i < 50; ++i) {
		Unique_Owner x(n);
		std::unique_ptr<int> p = x.values();
		std::this_thread::sleep_for(std::chrono::microseconds(10));
    }
}
  • Using the performance profiler in VS (Debug->Performance profiler)
  • the above code uses only 4MB
  • which means each iteration 4MB are allocated and then freed.

Figure 1

Templates

On many occasions we write multiple versions of the same code to handle different types. For example suppose we want to write a function to add two numbers (using the + operator) we write

int add(int x,int y){
    return x+y;
}
int add(double x,double y){
    return x+y;
}

Recall also that the + operator can be used to concatenate strings so we have to add that also. Since the all of those versions only the type changes, c++ allows us to pass the type as a parameters using templates.

#include <iostream>
#include <string>

template<typename T>
T add(T x,T y){
    return x+y;
}
int main(){
    int x=2,y=3;
    double u=3.4,v=3;
    std::string s="hello",k="there";

    std::cout<<add(x,y)<<std::endl;
    std::cout<<add(u,v)<<std::endl;
    std::cout<<add(s,k)<<std::endl;
}

In the above example the compiler automatically deduces the type which sometimes it cannot and we have to specify it as follows:

add<int>(x,y);
add<double>(u,v);
add<std::string>(s,k);
  • Note that the template is instantiated as needed at compile time. Also, we can pass parameters to the template other than types. For example
template <int n>
void doit(){
    int a[n];
}

Template specialization

  • The add example above doesn't make any sense if used with char.
  • Try adding the two characters 'a' and 'b'.
  • we would like to change the definition of add when the parameter type is char.
  • This is done using template specialization.
  • One reasonable "addition" of characters would be to obtain a character a the same distance from the last one.
  • For example, the distance (in ASCII code) between 'a' abd 'b' is 1 so the result of add('a','b')
    would be 'c' since b+1=c.

Below is the syntax for template specialization.

template<typename T>
T add(T x,T y){
   return x+y;
}
template<>
char add<char>(char a,char b){
  return a<b?b+(b-a):a+(a-b);
}
  • Note that the specialization starts with template<> and every occurence of T was replaced by char.
  • Even though the second implementation of add for char makes more sense that the first it is not satisfactory.
  • What we would like is for the result of add('a','b') to give "ab".
  • This means that the signature would become
    std::string add(char,char ).
  • We could be tempted to specialize it as follows
template<>
std::string add<char>(char a,char b){
    return std::string(1,a)+std::string(1,b);
}
  • But that is NOT a specialization.
  • The "general" version the return value is the same type as the input parameters which is not the case for our specialization.
  • The compiler will give an error.
  • You can check it here.
  • We can accomplish our aim by using auto as the return value, this way both versions will have the same signature.
template<typename T>
auto add(T a,T b){
   return a+b;
}
template<>
auto add<char>(char a,char b){
   //return std::string({a,b})//different way
   return std::string(1,a)+std::string(1,b);
}

You can try it here

Algorithms in STL

  • The standard template library STL defines a set of general purpose containers and algorithms.
  • You are almost always advised to use those instead of writing your own.
  • Furthermore, since C++17, most of the algorithms can take advantage of parallelisation.
  • In this section we look at a few examples.
  • Most of these algorithms need the <algorithm> or <numeric> header and they are defined over a range [start Iterator, end Iterator).

STL containers and iterators

  • Iterators are generalization of pointers
  • They present a common interface to all STL containers and algorithms.
  • For an array a pointer is sufficient since the elements of an array form a contiguous location in memory.
  • What if the elements are not stored contiguously?
  • Every container stores the elements differently,
  • it implements its own methods to iterate over its elements.
  • The user does not need to know the internal workings of the container
  • Given a container c an iterator itr points at an element stored in c.
  • Therefore dereferencing an iterator *itr will return the element itself.
  • Also iterators can be incremented and decremented like pointers: itr++ and itr--.
  • Furthermore, every container c has a begin and end() method.

Example with vector

#include <vector>
#include <iostream>
int main(){
  std::vector<std::string> sv;
  sv.push_back("one");
  sv.push_back("two");
  for(auto itr=sv.begin();itr!=sv.end();itr++){
      std::cout<<(*itr)<<std::endl;
  }
}
  • The auto keyword is useful since otherwise we have to write down the long type of the iterator:
  • std::vector<std::string>.

std::vector<std::string>::iterator itr;

  • vectors,unlike std::list, are required to store their content at contiguous locations.
  • Therefore they support random access to elements and thus define the index operator
std::vector<int> iv;
iv.push_back(1);
iv.push_back(2);
for(int i=0;i<iv.size();i++)
  iv[i]=i;
  • Since vectors are required by the c++ standard to use contiguous memory it is best to add and remove(as opposed to change) from the end of a vector.
  • While we will deal mostly with std::vector there are many other types of container/container adaptor in the STL
  • such as std::list, std::stack, std::map,etc.

Algorithms

  • In this section we explore a few algorithms provided by the STL.

  • std::count and std::count_if (signature)

#include <iostream>
#include <algorithm>
#include <vector>
bool is_even(int x){
    return x%2==0;
}
struct is_odd{
 bool operator()(int x){
     return x%2!=0;
 }
};

int main(){
    std::vector<int> v {7,2,4,2,2,8,2};
    int r=std::count(v.begin(),v.end(),2);
    std::cout<<" number of 2's is "<<r<<"\n";
    int even=std::count_if(v.begin(),v.end(),is_even);
    int odd=std::count_if(v.begin(),v.end(),is_odd());
    std::cout<<" number of even is "<<even<<" and odd is "<<odd<<"\n";
}

  • count_if takes a unary predicate as a third parameter and this can be either a function pointer or a function object.

  • Recall, a function object is one that defines an operator().

  • click here to run

  • std::find and std::find_if reference.
  • Find the first occurrence of an object in a container and returns an iterator pointing to it.
int main(){
  std::vector<int> v {3,7,9,1,1,8,19,2,1,4};
  auto first=std::find(v.begin(),v.end(),1);
  /* print all the elements from first
  * occurrence of the value 1 to the end 
  */
  for(auto itr=first;itr!=v.end();++itr)
    std::cout<<*itr<<",";
  std::cout<<"\n";
  first=std::find_if(v.begin(),v.end(),
            [](int x){return x%2==0;});
  /* all the elements from the first
   * occurrence of an even number to 
   * the end
   */
  for(auto itr=first;itr!=v.end();++itr)
    std::cout<<*itr<<",";
  std::cout<<"\n";
}

You can try it here

  • std::remove, std::remove_if and std::erase

  • The STL functions std::remove and std::remove_if do not remove anything.

  • They just partition the input range into two parts

  • a left part which contains the original objects, in the same order, where the desired object is removed,

  • a right part which contains "non useful" objects: either the one to be removed or additional copies of the already existing objects.

int main(){
    std::vector<int> v {1,2,3,4,2,5,2,6};
    auto new_end=std::remove(v.begin(),v.end(),2);
    for(auto itr=v.begin();itr!=new_end;++itr)
      std::cout<<*itr<<",";
    std::cout<<"\n";
    /* "non useful" part */
    for(auto itr=new_end;itr!=v.end();++itr)
     std::cout<<*itr<<",";
    std::cout<<"\n";
}
  • std::remove_if works the same way but using a predicate instead of a value.
  • These two functions are useful, especially, for the remove-erase idiom.
  • Once we have the "useful" range with the desired object removed from it we can actually erase it.
int main(){
    std::vector<int> v {1,2,3,4,2,5,2,6};
    auto new_end=std::remove(v.begin(),v.end(),2);
    v.erase(new_end,v.end());
    /* the vector contains only 
     * the "useful" part
     */
    for(auto x:v)
     std::cout<<x<<",";
    std::cout<<"\n";
}

You can try it here

  • Sometimes we need to print all the values in a container using a range-based for loop and auto.
  • We can type a little less if we defined an operator << that can handle containers.
template<typename Ostream,typename Container>
Ostream& operator<<(Ostream& os,Container& c) {
	for (auto x : c)
		os<< x << ",";
	return os;
}
  • The above works well with all sorts of containers, except with strings
  • In that case it will print it as characters separated by commas.
  • We can fix this by using template specialization
template <>
std::ostream& operator<< <std::ostream, std::string>
			(std::ostream& os, std::string& s) {
	/* cannot use os<<s because we enter infinite recursion */
	os.write(s.c_str(), s.size());
	return os;
}

substrings and subranges

  • The std::string class has a convenient member std::string::find
  • it returns the position of the first character of the substring in the string if found
  • and std::string::npos otherwise.
  • A general "substring" finding method is std::search which will find a "subrange" in any container, including strings.
int main(){
    std::string s {"First Middle Last Middle"};
    std::string sub {"Middle"};
    auto pos=s.find(sub);
    if(pos!=std::string::npos)std::cout<<"found at "<<pos<<"\n";
    else std::cout<<"not found\n";
    std::vector<int> v{4,2,7,3,1,5};
    std::vector<int> needle {7,3,1};
    /* returns an iterator to the beginning of the match
    * or v.end() if no match is found
    */
    auto itr=std::search(v.begin(),v.end(),needle.begin(),needle.end());
}

Lambda expressions

  • As we saw, many algorithms, especially in STL, take a callable object as a parameter.
  • Lambda expressions are a convenient way to define such a callable object.
  • We rewrite the previous example using lambda expressions.
int main(){
    std::vector<int> v {7,2,4,2,2,8,2};
    int r=std::count(v.begin(),v.end(),2);
    std::cout<<" number of 2's is "<<r<<"\n";
    int even=std::count_if(v.begin(),v.end(),
        [](int x){return x%2==0;});
    int odd=std::count_if(v.begin(),v.end(),
        [](int x){return x%2!=0;});
            std::cout<<" number of even is "<<even<<" and odd is "<<odd<<"\n";
    //counting the number of 2's in a different way
    // just to introduce capture
    int val=2;
    std::count_if(v.begin(),v.end(),
        [val](int x){return x==val;});
}

click here to run

  • The brackets "[]" introduces the lambda expression, sometimes called the capture list.

  • The "()" are the parameters passed to the expression, similar to function arguments.

  • Finally, between "{" and "}" is the body.

  • Note the capture of the variable val.

  • We could have used "[&val]" to capture it by reference.

  • If we want to capture all the variables we use "[=]" and by reference "[&]".