Skip to main content

Exit status

The smallest meaningful esque program: return a number from main and let the OS see it as the exit code.

The program​

# hello.esq
fn main() -> i32 = (10 + 3) * 2 - 1

Run​

$ ./esquec build hello.esq -o hello
$ ./hello; echo $?
25

What is happening​

  • fn main() -> i32 = … declares a zero-argument function returning i32. This is the entry point.
  • (10 + 3) * 2 - 1 is a single expression. It evaluates to 25.
  • The compiler emits a tiny _start that calls main, then issues the Linux exit_group syscall with main's return value.

Try​

  • Replace the body with 42. Re-run. echo $? → 42.
  • Replace it with 'A'. (Char literals are i32 codepoints.) → 65.
  • Replace it with 99_i32. (Explicit type suffix.) → 99.
  • Replace it with 1 + 1.0. The compiler refuses to compile it.

Inspect the build​

$ ./esquec build hello.esq --emit=ceir
fn main() -> i32 {
  %0 = ConstInt 13
  %1 = ConstInt 2
  %2 = Mul %0 %1
  %3 = ConstInt 1
  %4 = Sub %2 %3
  Return %4
}

13 and 26 - 1 have already been computed at the AST level by the constant-fold pass.

$ ./esquec build hello.esq --emit=asm
main:
  b8 19 00 00 00       mov eax, 0x19      ; 25
  c3                   ret

The whole program is two instructions.