Use of explicit keyword in C++
Last Updated :
14 Sep, 2022
Explicit Keyword in C++ is used to mark constructors to not implicitly convert types in C++. It is optional for constructors that take exactly one argument and work on constructors(with a single argument) since those are the only constructors that can be used in typecasting.
Let’s understand explicit keyword through an example.
Predict the output of the following C++ Program
CPP
#include <iostream>
using namespace std;
class Complex {
private :
double real;
double imag;
public :
Complex( double r = 0.0,
double i = 0.0) : real(r),
imag(i)
{
}
bool operator == (Complex rhs)
{
return (real == rhs.real &&
imag == rhs.imag);
}
};
int main()
{
Complex com1(3.0, 0.0);
if (com1 == 3.0)
cout << "Same" ;
else
cout << "Not Same" ;
return 0;
}
|
As discussed in this article, in C++, if a class has a constructor which can be called with a single argument, then this constructor becomes a conversion constructor because such a constructor allows conversion of the single argument to the class being constructed. In this case, when com1 == 3.0 is called, 3.0 is implicitly converted to Complex type because the default constructor can be called with only 1 argument because both parameters are default arguments and we can choose not to provide them.
We can avoid such implicit conversions as these may lead to unexpected results. We can make the constructor explicit with the help of an explicit keyword. For example, if we try the following program that uses explicit keywords with a constructor, we get a compilation error.
CPP
#include <iostream>
using namespace std;
class Complex {
private :
double real;
double imag;
public :
explicit Complex( double r = 0.0,
double i = 0.0) :
real(r), imag(i)
{
}
bool operator == (Complex rhs)
{
return (real == rhs.real &&
imag == rhs.imag);
}
};
int main()
{
Complex com1(3.0, 0.0);
if (com1 == 3.0)
cout << "Same" ;
else
cout << "Not Same" ;
return 0;
}
|
Output
Compiler Error : no match for 'operator==' in 'com1 == 3.0e+0'
We receive an error here because to avoid any unexpected errors we have made our constructor an explicit constructor and 3.0 won’t be converted to Complex by our constructor on its own.
We can still typecast the double values to Complex, but now we have to explicitly typecast it. For example, the following program works fine.
CPP
#include <iostream>
using namespace std;
class Complex {
private :
double real;
double imag;
public :
explicit Complex( double r = 0.0,
double i = 0.0):
real(r) , imag(i)
{
}
bool operator == (Complex rhs)
{
return (real == rhs.real &&
imag == rhs.imag);
}
};
int main()
{
Complex com1(3.0, 0.0);
if (com1 == (Complex)3.0)
cout << "Same" ;
else
cout << "Not Same" ;
return 0;
}
|
Note: The explicit specifier can be used with a constant expression. However, if that constant expression evaluates to true, then only the function is explicit.
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