This code is complied without warnings: fn main() { let mut _v : Vec<f64> = vec![]; let mut count : f64 = 0.; let _median : f64 = { let mut sum :f64 = 0.; for i in &_v { sum += *i; count += 1.; println!("never call"); } …

NaN is a valid floating point special value.

It looks like the defect in Rust compiler. Because there is an announcement that Rust compiler should catch the situation of uninitialized variable. So if Nan is valid why there is that valid value in uninitialized variable?

NaN is not the same as the value being uninitialized. The value NaN is an actual specific floating point value like 2.0, 7.5 or infinity.

If you want an example of an uninitialized value, here you go: fn main() { let mut val: f64; if false { val = 10.0; } println!("{}", val); } error[E0381]: borrow of possibly-uninitialized variable: `val` --> src/main.rs:9:20 | 9 | println!("{}", val); | …

So the question why is Nan in the _median? There is no assigning Nan to the _median.

Why doesn't Rust throw an exception like in the situation of i32? A signaling NaN in any arithmetic operation (including numerical comparisons) will cause an "invalid operation.

Most languages generate a NaN value when you divide by zero. This is not Rust being weird, if anything, it's floating point being weird.

I'll be reducing your example : fn main() { let mut _v : Vec<f64> = vec![]; let mut count : f64 = 0.; let _median : f64 = { let mut sum :f64 = 0.; for i in &_v { sum += *i; count += 1.; println!("never call"); } sum …

Thank you. Could you please send the link to the specs? Actually Rust is going to be the language that helps to prevent various sort of errors. And idea is very good! But in the real situation (my testcase is about it) Rust doesn't help to catch an error. It's bad.

Getting Nan value in safe code

code review

alice May 28, 2021, 2:19pm 8

It's because you are dividing by zero. You are doing this because when the list is empty, this results in count being zero.

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Getting Nan value in safe code

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