CODEWITHSUNDEEP
c++syntax
ajith.mk
ajith.mk

Reputation: 85

Casting to template type

What is this syntax Prof. Stroustrup uses in his book The C++ Programming Language 4th ed. while describing hash and equality functions on page 917.

std::hash<int>{}(variable) //variable is of type int

Is that a casting from int to hash ? But why those flower brackets after hash? I know that placing them after a variable default initializes it. As for as casting is concerned we normally cast like say double (int)!

Upvotes: 3

Views: 651

Answers (3)

Some programmer dude
Some programmer dude

Reputation: 409136

Lets break std::hash<int>{}(variable) down into its components:

  • std::hash<int> - This is the type, it's a specific type of standard hash template.

  • {} - This creates an instance of the std::hash<int> class.

  • (variable) - This calls the function call operator on the instance previously created, passing variable as argument.

After the expression, the instance of the std::hash<int> object is destructed.

For example:

std::size_t hash = std::hash<int>{}(variable);

is roughly equivalent to

std::size_t hash;
{
    std::hash<int> hashing_temporary_object;
    hash = hashing_temporary_object(variable);
    // The above call is equal to hashing_temporary_object.operator()(variable)
}

Upvotes: 8

aschepler
aschepler

Reputation: 72271

std::hash is a class template. <int> provides template arguments to the template. std::hash<int> together is the type produced by a specialization of the class template. std::hash<int>{} default-constructs a value of that type. std::hash<int>{}(variable) calls the class's operator() to evaluate the hash for a given number.

Upvotes: 1

doctorlove
doctorlove

Reputation: 19232

std::hash<int>{} creates a function object. As you say the "flowery" or even curly braces initialise this.

You then call the function with the int.

Upvotes: 0

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