Programming

C++ How to make cout not use scientific notation

 in C++  tagged C++ / cout / fixed / I/O manipulator / scientific notation / std::fixed by

To force cout to print numbers exactly as they are and prevent it from using the scientific notation, we can use the std::fixed I/O manipulator as follows

#include <iostream>

using namespace std;

int main()
{
    std::cout << "The number 0.0001 in fixed:      " << std::fixed << 0.0001 << endl
              << "The number 0.0001 in default:    " << std::defaultfloat << 0.0001 << endl;

    std::cout << "The number 1000000000.0 in fixed:      " << std::fixed << 1000000000.0 << endl
              << "The number 1000000000.0 in default:    " << std::defaultfloat << 1000000000.0 << endl;
return 0;
}

Output

The number 0.0001 in fixed:      0.000100
The number 0.0001 in default:    0.0001
The number 1000000000.0 in fixed:      1000000000.000000
The number 1000000000.0 in default:    1e+09

C++: Simplified version of ‘Friends with benefits’ demonstrating friend classes

 in C++ / Funny  tagged friend / friend class / private by

A friend class in C++ can access the private and protected members of the class in which it is declared as a friend.

Friendship may allow a class to be better encapsulated by granting per-class access to parts of its API that would otherwise have to be public.[2] This increased encapsulation comes at the cost of tighter coupling due to interdependency between the classes.

Properties

  • Friendships are not symmetric – if class A is a friend of class B, class B is not automatically a friend of class A.
  • Friendships are not transitive – if class A is a friend of class B, and class B is a friend of class C, class A is not automatically a friend of class C.
  • Friendships are not inherited – if class Base is a friend of class X, subclass Derived is not automatically a friend of class X; and if class X is a friend of class Base, class X is not automatically a friend of subclass Derived.

From Wikipedia: https://en.wikipedia.org/wiki/Friend_class

In the following example we assign both the Man to be a friend of the Woman and the Woman to be a friend of the Man in order to allow both parties to access the private members of the other.

#include <iostream>
using namespace std;

class Man;

class Woman {
  friend class Man;

public:
  void touch(Man man);
private:
  void * body;
};

class Man {
  friend class Woman;

public:
  void touch(Woman woman);
private:
  void * body;
};

void Woman::touch(Man man) {
  void * other = man.body;
}

void Man::touch(Woman woman) {
  void * other = woman.body;
}

int main() {
  Man man;
  Woman woman;

  man.touch(woman);
  woman.touch(man);
  return 0;
}

C++: “undefined reference to” templated class function 6

 in C++  tagged C++ / template / undefined reference to by

In case you have a project where you use a templated class that is split in its own header (.h) and source (.cpp) files, if you compile the class, into an object file (.o), separately from the code that uses it, you will get the undefined reference to error at linking.

Lets assume we have Stack.cpp and Stack.h which define a templated stack using vectors. And main.cpp that uses this class after including Stack.h.

If you try to compile these files as mentioned above, one by one, later you will get a linking error saying undefined reference to for the methods of the class.

The code in the template is not sufficient to instruct the compiler to produce the methods that are needed by main.cpp (e.g. Stack<int>::push(...) and Stack<string>::push(...)) as the compiler does not know, while compiling Stack.cpp by itself, the data types it should provide support for.

The reason it allows you to compile these incomplete objects is the following:

  • main.cpp: the compiler will implicitly instantiate the template classes Stack<int> and Stack<string> because those particular instantiations are requested in main.cpp. Since the implementations of those member functions are not in main.cpp, nor in any header file included in main.cpp (particularly Stack.h), the compiler will not include complete versions of those functions in main.o and it will expect to find them in another object during linking.
  • Stack.cpp: the compiler won’t compile the instantiations of Stack<int> and Stack<string> neither as there are no implicit or explicit instantiations of them in Stack.cpp nor Stack.h.

So in the end, neither of the .o files contain the actual implementations of Stack<int> and Stack<string> and the linking fails.

Solutions

Solution 1 : Explicitly instantiate the template

At the end of Stack.cpp, you can explicitly instantiate all needed templates.
In our example we would add:


template class Stack<int>;
template class Stack<std::string>;

This will ensure that, when the compiler is compiling Stack.cpp that it will explicitly compile all the code needed for the Stack<int> and Stack<std::string> classes.

Using this method, you should ensure that all the of the implementation is placed into one .cpp file and that the explicit instantation is placed after the definition of all the functions (for example, at the end of the file).

A problem with this method is that it forces you to update the Stack.cpp file each time you want to add support for a new data type (or remove one).

Solution 2 : Move the implementation code into the header file

Move all the source code of Stack.cpp to Stack.h, and then delete Stack.cpp. Using this method you do not need to manually instantiate all possible data types that are needed and thus you do not need to modify code of the class. As a side-effect, if you use the header file in many other source files, it will compile the functions of the header file in each source. This can make compilation slower but it will not create any compilation/linking problems, as the linker will ignore the duplicate implementations.

Solution 3 : Move the implementation code into a new header file and include it in the original header file

Rename Stack.cpp to Stack_impl.h, and then include Stack_impl.h from Stack.h to keep the implementation in a separate file from the declaration. This method will behave exactly like Solution 2.

How does the expression, “*pointer++” evaluate?

 in C / C++  tagged C/C++ / increment / pointer / pointers by

*pointer++ will increment the position of the pointer first but it will return the value that was pointed before the position of the pointer changed.

*pointer++ is equivalent to *(pointer++). This happens because the postfix ++ and -- operators have higher precedence than the indirection (dereference).
You can read more about the precedence order at this very helpful article at Wikipedia: https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Operator_precedence

To increment the value pointed to by pointer, use (*pointer)++.

To increment the position of the pointer and return the value that is pointed after the position of the pointer changed use *++pointer.

Examples

The following example will increment the position of the pointer but return the value that was originally pointed.
By the end of the following block, pointer will point to position 1 and the value variable will have the value 11.


const unsigned int values[] = {11, 12, 14, 18};
const unsigned int *pointer = values;
const unsigned int value = *pointer++;

The following example will increment the position of the pointer and return the value that the new position is pointing to.
By the end of the following block, pointer will point to position 1 and the value variable will have the value 12.


const unsigned int values[] = {11, 12, 14, 18};
const unsigned int *pointer = values;
const unsigned int value = *++pointer;