Directly calling `"msp430-interrupt" fn` should also be invalid

[workingjubilee]

I tried this code:

#![feature(abi_msp430_interrupt)]

pub extern "msp430-interrupt" fn msp430_isr() {}

pub fn call() {
    msp430_isr();
}

I expected to see rustc reject the erroneous code somewhere before final LLVM lowering, because the notion of "calling" an interrupt still seems nonsensical to me, no matter what its signature is.

Instead, this happened:

rustc-LLVM ERROR: ISRs cannot be called directly

Meta

rustc --version --verbose:

rustc 1.84.0-nightly (a0d98ff0e 2024-10-31)
binary: rustc
commit-hash: a0d98ff0e5b6e1f2c63fd26f68484792621b235c
commit-date: 2024-10-31
host: x86_64-unknown-linux-gnu
release: 1.84.0-nightly
LLVM version: 19.1.1

@rustbot label: +A-hardware-interrupts +A-LLVM +O-msp430 +A-ABI +T-compiler

Related Issues

• Tracking issue for the msp430-interrupt calling convention/ABI #38487
• Directly calling "x86-interrupt" fn should be invalid #132834
• Directly calling extern "riscv-interrupt-{m,s}" fn... compiles? #132836

[GrigorenkoPV]

Fixed by #142597