When code is compiled using the X87 FPU, comparing to infinity can lead to unexpected behavior.

I tried this code:

#[inline(never)]
fn get_num() -> f64 {
    let num: f64 = 1.0e300;
    // volatile is to avoid optimizations
    unsafe { std::ptr::read_volatile(&num) }
}

fn main() {
    let x = get_num();
    let y = get_num();
    let z = x * y;
    if z != f64::INFINITY && z != f64::NEG_INFINITY && !z.is_nan() {
        let exp = (z.to_bits() >> 52) & 0x7FF;
        assert!(exp != 0x7FF);
        println!("is finite, exp = {}", exp);
    } else {
        println!("is not finite");
    }
}

I expected to see this happen:

This should either print "infinite or nan" or "is finite, exp = ...". The assert! should never fail because the exponent bits are 0x7FF if and only if the number is infinity or nan, so the if condition should have been false.

Instead, this happened:

When compiled for X86 without SSE, the assert will fail. This could also cause safety issues if unsafe code relies on it.

The following command can be used to compile without SSE:

RUSTFLAGS="-C target-cpu=pentium" cargo run --target i686-unknown-linux-gnu

Meta

rustc --version --verbose:

rustc 1.43.1 (8d69840ab 2020-05-04)
binary: rustc
commit-hash: 8d69840ab92ea7f4d323420088dd8c9775f180cd
commit-date: 2020-05-04
host: x86_64-unknown-linux-gnu
release: 1.43.1
LLVM version: 9.0

and

rustc 1.45.0-nightly (a74d1862d 2020-05-14)
binary: rustc
commit-hash: a74d1862d4d87a56244958416fd05976c58ca1a8
commit-date: 2020-05-14
host: x86_64-unknown-linux-gnu
release: 1.45.0-nightly
LLVM version: 9.0