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Home/ Questions/Q 743709
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T08:53:41+00:00 2026-05-14T08:53:41+00:00

How does ML perform the type inference in the following function definition: let add

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How does ML perform the type inference in the following function definition:

let add a b = a + b

Is it like C++ templates where no type-checking is performed until the point of template instantiation after which if the type supports the necessary operations, the function works or else a compilation error is thrown ?

i.e. for example, the following function template

template <typename NumType>
NumType add(NumType a, NumType b) {
  return a + b;
}

will work for

add<int>(23, 11);

but won’t work for

add<ostream>(cout, fout);

Is what I am guessing is correct or ML type inference works differently?

PS: Sorry for my poor English; it’s not my native language.

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  1. Editorial Team
    Editorial Team
    2026-05-14T08:53:42+00:00Added an answer on May 14, 2026 at 8:53 am

    I suggest you have a look at this article: What is Hindley-Milner? (and why is it cool)

    Here is the simplest example they use to explain type inference (it’s not ML, but the idea is the same):

    def foo(s: String) = s.length
    // note: no explicit types
    def bar(x, y) = foo(x) + y
    

    Just looking at the definition of bar, we can easily see that its type must be (String, Int)=>Int. That’s type inference in a nutshell. Read the whole article for more information and examples.

    I’m not a C++ expert, but I think templates are something else that is closer to genericity/parametricity, which is something different.

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