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Due to dead code elimination, the compiler already omits all of that part of stdlib that your code doesn’t use.

by nequo 3y ago

pornel 3y ago
Not quite. Every Rust program will have some code path that may panic, and the default panic handler uses debug formatting, which uses dynamic dispatch, which prevents elimination of the rest of the printing machinery. There’s panic_immediate_abort unstable setting that makes Rust panics crash as hard as a C segfault, and only then you can get rid of a good chunk of stdlib.
- nequo 3y ago
  Does this mean that only the printing machinery is not eliminated or that other parts of stdlib are present in the binary too even though unused?
  - fl0ki 3y ago
    The printing machinery alone is quite large when you consider that it includes the code & raw data for Unicode, whether or not similar facilities were already available on the host libc. Though you're not likely to avoid that in any non-trivial Rust program anyway, as even a pretty barebones CLI will need Unicode-aware string processing. I generally find Rust binaries to be "a few" megabytes if they don't have an async runtime, and a few more if they do. It has never bothered me on an individual program basis, but I can imagine it adding up over an entire distribution with hundreds of individual binaries. I see the very real concern there, but personally I would still not risk ABI hazards just to save on space.
- bpye 3y ago
  The printing machinery is quite unfortunate. Beyond being large, dynamic dispatch makes any attempt at stack size analysis much harder. I’ve used Rust for some embedded side projects and I really wish there was a way to just get some unique identifier that I could translate (using debug symbols) to a filename and line number for a crash. This would sort of be possible if you could get the compiler to put the filenames in a different binary section, as you could then just save the address of the string and strip out the actual strings - but today that’s not possible.

Due to dead code elimination, the compiler already omits all of that part of stdlib that your code doesn’t use.

by nequo 3y ago

Due to dead code elimination, the compiler already omits all of that part of stdlib that your code doesn’t use.

Not quite. Every Rust program will have some code path that may panic, and the default panic handler uses debug formatting, which uses dynamic dispatch, which prevents elimination of the rest of the printing machinery. There’s panic_immediate_abort unstable setting that makes Rust panics crash as hard as a C segfault, and only then you can get rid of a good chunk of stdlib.

nequo 3y ago
Does this mean that only the printing machinery is not eliminated or that other parts of stdlib are present in the binary too even though unused?
- fl0ki 3y ago
  The printing machinery alone is quite large when you consider that it includes the code & raw data for Unicode, whether or not similar facilities were already available on the host libc. Though you're not likely to avoid that in any non-trivial Rust program anyway, as even a pretty barebones CLI will need Unicode-aware string processing. I generally find Rust binaries to be "a few" megabytes if they don't have an async runtime, and a few more if they do. It has never bothered me on an individual program basis, but I can imagine it adding up over an entire distribution with hundreds of individual binaries. I see the very real concern there, but personally I would still not risk ABI hazards just to save on space.
bpye 3y ago
The printing machinery is quite unfortunate. Beyond being large, dynamic dispatch makes any attempt at stack size analysis much harder. I’ve used Rust for some embedded side projects and I really wish there was a way to just get some unique identifier that I could translate (using debug symbols) to a filename and line number for a crash. This would sort of be possible if you could get the compiler to put the filenames in a different binary section, as you could then just save the address of the string and strip out the actual strings - but today that’s not possible.