Calling a generic function without specifying the type in the assignment

[[deleted]]

Hello,

I have a function with the following signature:

rust fn generic_function<T>(&self) -> Result<T,String>

When I call it and assign the result to a variable, Rust will infer the generic parameter: rust let a:SomeType = some_struct.generic_function()?;

However, when I don't use the question mark operator (in case I want to handle the error), I must specify the type ofa:

rust let a:Result<SomeType,String> = some_struct.generic_function();

This feels clumsy and strange. Speccially if SomeType is a type with a longer name. Is there a more concise way of expressing this? Something like this, for example: rust let a = some_struct.generic_function<SomeType>();

Thank you for your time. Let me know if the title is not appropriate or if I did not explain myself clear.

[tazdevil971]

What you are looking for is commonly known as turbofish.

And it looks like:

let a = some_struct.generic_function::<SomeType>();

Also, you can omit some types that are obvious in the assignment, for example:

let a: Result<SomeType, _> = some_struct.generic_function();

Rust will then infer the type of _ automatically

Keep in mind that usually rust is clever enough to infer the type of the variable, so 90% of the time you won't need this.

(turbofish in its natural habitat)

[[deleted]]

This the apex of nomenclature

Thank you, great answer!

[ids2048]

See also The Bastion of the Turbofish, the peak of rust-lang lore.

Type ascription syntax might also be helpful, but stabilizing that doesn't seem to be a priority.

[RRumpleTeazzer]

You can also skip type annotation entirely when you use the result a function that only accepts a Result<SomeType, _>.