Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is 'var' weak type.

  1. Jan 21, 2007 #1
    As far as I understand type systems in strongly typed system such as the one found in Java you must specify the exact type of the variable. Like e.g.

    int x = 5; // strongly typed

    also this would fail

    Object x = 5;
    Object y = "Not soo strong are we?";
    Object z = x + y; // would throw a compiler error, since + operation is not defined on Object + Object, it is only defined on either Number + Number or String + String or String + Object

    Now I said to a friend that C# is no longer strongly typed since in 3.0 you have the var variable which is more of a placeholder than a type, what kind of type is var? var can substitue any type therefore I would say that is weak type. He didn't agree on that one. I am left a bit puzzled. Is var type weak or not? I would say it is.
  2. jcsd
  3. Jan 21, 2007 #2


    User Avatar
    Homework Helper

    It's still strongly typed because the compiler infers the actual type at compile time. In no case would it need a runtime check; I think in ambiguous cases the type must be supplied. It's just a matter of convenience I guess, but I don't understand why it is seen as a good thing. To me it seems that it would make code more difficult to understand.
  4. Jan 21, 2007 #3
    Thanks for the answer. Yeah, I wouldn't dream of using var exept if I would write JavaScript.
  5. Jan 21, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    var is a feature that would be very useful for template programming in C++.

    Type expressions can get quite verbose; here's a simple one:

    Code (Text):

    // Print everything in a container
    template <typename Container>
    void print_all(const Container &c) {
      typename Container::const_iterator iter = c.begin();
      while(iter != c.end()) {
        std::cout << (*iter++) << std::endl;
    Sometimes, you have to write stuff like typename Container::const_iterator a lot; it would be very convenient to be able to just write:

    var iter = c.begin()

    Even worse, sometimes you aren't even supposed to know the return type of an expression. For example, in linear algebra packages, if I write an expression like:

    Code (Text):

    template <typename T>
    void do_something(boost::numeric::ublas::vector<T> &a,
             const boost::numeric::ublas::vector<T> &b,
             const boost::numeric::ublas::vector<T> &c)
      a = 3 * b + 4 * c;
    ublas generates efficient code for this function through the magic of expression templates: each of those arithmetic operations constructs a type that "knows" the parse tree of the expression. For example, the return type of 3*b is

    typename vector_binary_scalar1_traits<int, vector<T>, scalar_multiplies<int, typename vector<T>::value_type> >::result_type

    which is not only something you wouldn't want to have to type, it's something that you really ought not to even know about! If I wanted to do something like

    Code (Text):

    var e1 = 3 * b;
    var e2 = 4 * c;
    a = e1 + e2;
    then a keyword like var is the only reasonable option for doing so.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Is 'var' weak type.