TypeScript Language Specification v1.5

An example:
signature are fixed, and e's type is changed to a function type with e's call signature instantiated
in the context of T's call signature (section 3.10.5).
function choose(x: T, y: T): T {
return Math.random() < 0.5 ? x : y;
}
var x = choose(10, 20); // Ok, x of type number
var y = choose("Five", 5); // Error
In the first call to 'choose', two inferences are made from 'number' to 'T', one for each parameter. Thus,
'number' is inferred for 'T' and the call is equivalent to
var x = choose(10, 20);
In the second call to 'choose', an inference is made from type 'string' to 'T' for the first parameter and an
inference is made from type 'number' to 'T' for the second parameter. Since neither 'string' nor 'number' is
a supertype of the other, type inference fails. That in turn means there are no applicable signatures and
the function call is an error.
In the example
function map(a: T[], f: (x: T) => U): U[] {
var result: U[] = [];
for (var i = 0; i < a.length; i++) result.push(f(a[i]));
return result;
}
var names = ["Peter", "Paul", "Mary"];
var lengths = map(names, s => s.length);
inferences for 'T' and 'U' in the call to 'map' are made as follows: For the first parameter, inferences are
made from the type 'string[]' (the type of 'names') to the type 'T[]', inferring 'string' for 'T'. For the second
parameter, inferential typing of the arrow expression 's => s.length' causes 'T' to become fixed such that
the inferred type 'string' can be used for the parameter 's'. The return type of the arrow expression can
then be determined, and inferences are made from the type '(s: string) => number' to the type '(x: T) =>
U', inferring 'number' for 'U'. Thus the call to 'map' is equivalent to
var lengths = map(names, s => s.length);
and the resulting type of 'lengths' is therefore 'number[]'.
In the example
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