min-sized-rust-windows

I completely forgot stdout handle is in the PEB! Thanks! After learning how the windows x64 syscall calling convention works I got it working on win10. https://i.imgur.com/ErHU7Kr.png https://github.com/mcountryman/min-sized-rust-windows/commit...

It's very easy and fast to try: using 0.7.0+51d7c14ce from https://ziglang.org/download/ and: const std = @import("std"); pub fn main() void { std.debug.print("Hello, world!\n", .{}); } ---------- zig build-exe hello.zig -O ReleaseSmall --strip --single-threaded -target x86_64-windows Resulting hello.exe is 3072 bytes. From these 2560 are zero bytes. The only system calls from the resulting binary are to ExitProcess, GetLastError and WriteFile from KERNEL32.dll Additionally, changing the release optimizations: zig build-exe hello.zig -O ReleaseFast --single-threaded --strip -target x86_64-windows produces hello.exe of 2560 bytes, 1849 from which are zeroes. Note all: winter_blue has right to mention Zig. It's about what's provided out of the box in that new language and the provided optimizations are really effective.

I notice you moved from asm! to llvm_asm! [0]. Since the commit message doesn't explain why, would you be able to do so here? (Purely for my own curiosity) [0] https://github.com/mcountryman/min-sized-rust-windows/commit...

I'm disappointed. In C I can get a 1k Hello Word binary simply be writing int main(void) { puts("Hello World!"); return 0; } The Rust version is a mess [0] [0] https://github.com/mcountryman/min-sized-rust-windows/blob/m...