Tuesday 16 January 2018

c++ - What does the explicit keyword mean?

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What does the
explicit keyword mean in C++?



Answer




The compiler is allowed to make one implicit
conversion to resolve the parameters to a function. What this means is that the compiler
can use constructors callable with a single parameter to
convert from one type to another in order to get the right type for a parameter.




Here's an example class with a
constructor that can be used for implicit
conversions:



class
Foo
{
public:
// single parameter constructor, can be
used as an implicit conversion
Foo (int foo) : m_foo (foo)

{
}


int GetFoo () { return m_foo;
}

private:
int
m_foo;
};


Here's
a simple function that takes a Foo
object:



void DoBar (Foo
foo)

{
int i = foo.GetFoo
();
}


and
here's where the DoBar function is
called.



int main
()
{
DoBar
(42);

}


The
argument is not a Foo object, but an
int. However, there exists a constructor for
Foo that takes an int so this
constructor can be used to convert the parameter to the correct
type.



The compiler is allowed to do this once
for each parameter.



Prefixing the
explicit keyword to the constructor prevents the compiler from
using that constructor for implicit conversions. Adding it to the above class will
create a compiler error at the function call DoBar (42). It is
now necessary to call for conversion explicitly with DoBar (Foo
(42))



The reason you might want to
do this is to avoid accidental construction that can hide bugs. Contrived
example:





  • You
    have a MyString(int size) class with a constructor that
    constructs a string of the given size. You have a function print(const
    MyString&)
    , and you call print(3) (when you
    actually intended to call print("3")). You
    expect it to print "3", but it prints an empty string of length 3
    instead.


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