Question

What does the explicit keyword mean in C++?

Answer 1

The compiler is allowed to make one implicit conversion to resolve the parameters to a function. This means 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 with converting constructors that shows how it works:

struct Foo {     // Single parameter constructor, can be used as an implicit conversion.     // Such a constructor is called "converting constructor".     Foo(int x) {} }; struct Faz {     // Also a converting constructor.     Faz(Foo foo) {} };  // The parameter is of type Foo, not of type int, so it looks like // we have to pass a Foo. void bar(Foo foo);  int main() {     // However, the converting constructor allows us to pass an int.     bar(42);     // Also allowed thanks to the converting constructor.     Foo foo = 42;     // Error! This would require two conversions (int -> Foo -> Faz).     Faz faz = 42; } 

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 bar(42). It is now necessary to call for conversion explicitly with bar(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 class with a constructor that constructs a string of the given size. You have a function print(const MyString&) (as well as an overload print (char *string)), 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.