[dragonwriter]

> Same reason we don't have to type `(((2int + 2int)int) / 8int)int`. Type inference already exists in any ALGOL-like.

That's not why you don't have to do that.

Many (most statically typed?) Algol-likes have strict types for literals (e.g., 2 is always a signed int, if you if you want specifically unsigned you might say 2u and if you want a double you say 2.0, and if you want a single-precision float you say 2.0f, or something) and strict rules at how math between them works and what types it produces, this has been true since long before tyoe inference became common, and is why you don't have to say 2int+2int — 2 is syntactically defined as int.

There is no inference

[goto11]

It is still type inference though. If 2 is an int, the compiler perform type inference to figure out that the expression 2 + 2 is also an int. It is just that traditional languages only use type inference for expressions, not declarations.

[dllthomas]

It's type propagation. Which can be seen as a form of inference or not, depending just where you draw the line.

[gpderetta]

So is auto, decltype and templates. In C++ we properly call it type deduction to distinguish it from actual H-M style inference which C++ lacks. The details of how it is called doesn't detract from parent's argument.

[dllthomas]

Yes. I hadn't meant my comment as argument on either side, just added context.

[gpderetta]

What's the type of the binary + operator in C?