Why does “noreturn” function return?

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执念已碎
执念已碎 2020-12-25 10:51

I read this question about noreturn attribute, which is used for functions that don\'t return to the caller.

Then I have made a program in C.

         


        
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  • 2020-12-25 11:15

    TL:DR: It's a missed-optimization by gcc.


    noreturn is a promise to the compiler that the function won't return. This allows optimizations, and is useful especially in cases where it's hard for the compiler to prove that a loop won't ever exit, or otherwise prove there's no path through a function that returns.

    GCC already optimizes main to fall off the end of the function if func() returns, even with the default -O0 (minimum optimization level) that it looks like you used.

    The output for func() itself could be considered a missed optimization; it could just omit everything after the function call (since having the call not return is the only way the function itself can be noreturn). It's not a great example since printf is a standard C function that is known to return normally (unless you setvbuf to give stdout a buffer that will segfault?)

    Lets use a different function that the compiler doesn't know about.

    void ext(void);
    
    //static
    int foo;
    
    _Noreturn void func(int *p, int a) {
        ext();
        *p = a;     // using function args after a function call
        foo = 1;    // requires save/restore of registers
    }
    
    void bar() {
            func(&foo, 3);
    }
    

    (Code + x86-64 asm on the Godbolt compiler explorer.)

    gcc7.2 output for bar() is interesting. It inlines func(), and eliminates the foo=3 dead store, leaving just:

    bar:
        sub     rsp, 8    ## align the stack
        call    ext
        mov     DWORD PTR foo[rip], 1
       ## fall off the end
    

    Gcc still assumes that ext() is going to return, otherwise it could have just tail-called ext() with jmp ext. But gcc doesn't tailcall noreturn functions, because that loses backtrace info for things like abort(). Apparently inlining them is ok, though.

    Gcc could have optimized by omitting the mov store after the call as well. If ext returns, the program is hosed, so there's no point generating any of that code. Clang does make that optimization in bar() / main().


    func itself is more interesting, and a bigger missed optimization.

    gcc and clang both emit nearly the same thing:

    func:
        push    rbp            # save some call-preserved regs
        push    rbx
        mov     ebp, esi       # save function args for after ext()
        mov     rbx, rdi
        sub     rsp, 8          # align the stack before a call
        call    ext
        mov     DWORD PTR [rbx], ebp     #  *p = a;
        mov     DWORD PTR foo[rip], 1    #  foo = 1
        add     rsp, 8
        pop     rbx            # restore call-preserved regs
        pop     rbp
        ret
    

    This function could assume that it doesn't return, and use rbx and rbp without saving/restoring them.

    Gcc for ARM32 actually does that, but still emits instructions to return otherwise cleanly. So a noreturn function that does actually return on ARM32 will break the ABI and cause hard-to-debug problems in the caller or later. (Undefined behaviour allows this, but it's at least a quality-of-implementation problem: https://gcc.gnu.org/bugzilla/show_bug.cgi?id=82158.)

    This is a useful optimization in cases where gcc can't prove whether a function does or doesn't return. (It's obviously harmful when the function does simply return, though. Gcc warns when it's sure a noreturn function does return.) Other gcc target architectures don't do this; that's also a missed optimization.

    But gcc doesn't go far enough: optimizing away the return instruction as well (or replacing it with an illegal instruction) would save code size and guarantee noisy failure instead of silent corruption.

    And if you're going to optimize away the ret, optimizing away everything that's only needed if the function will return makes sense.

    Thus, func() could be compiled to:

        sub     rsp, 8
        call    ext
        # *p = a;  and so on assumed to never happen
        ud2                 # optional: illegal insn instead of fall-through
    

    Every other instruction present is a missed optimization. If ext is declared noreturn, that's exactly what we get.

    Any basic block that ends with a return could be assumed to never be reached.

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  • 2020-12-25 11:21

    As others have mentioned, this is classic undefined behavior. You promised func wouldn't return, but you made it return anyway. You get to pick up the pieces when that breaks.

    Although the compiler compiles func in the usual manner (despite your noreturn), the noreturn affects calling functions.

    You can see this in the assembly listing: the compiler has assumed, in main, that func won't return. Therefore, it literally deleted all of the code after the call func (see for yourself at https://godbolt.org/g/8hW6ZR). The assembly listing isn't truncated, it literally just ends after the call func because the compiler assumes any code after that would be unreachable. So, when func actually does return, main is going to start executing whatever crap follows the main function - be it padding, immediate constants, or a sea of 00 bytes. Again - very much undefined behavior.

    This is transitive - a function that calls a noreturn function in all possible code paths can, itself, be assumed to be noreturn.

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  • 2020-12-25 11:23

    noreturn is a promise. You're telling the compiler, "It may or may not be obvious, but I know, based on the way I wrote the code, that this function will never return." That way, the compiler can avoid setting up the mechanisms that would allow the function to return properly. Leaving out those mechanisms might allow the compiler to generate more efficient code.

    How can a function not return? One example would be if it called exit() instead.

    But if you promise the compiler that your function won't return, and the compiler doesn't arrange for it to be possible for the function to return properly, and then you go and write a function that does return, what's the compiler supposed to do? It basically has three possibilities:

    1. Be "nice" to you and figure out a way to have the function return properly anyway.
    2. Emit code that, when the function improperly returns, it crashes or behaves in arbitrarily unpredictable ways.
    3. Give you a warning or error message pointing out that you broke your promise.

    The compiler might do 1, 2, 3, or some combination.

    If this sounds like undefined behavior, that's because it is.

    The bottom line, in programming as in real life, is: Don't make promises you can't keep. Someone else might have made decisions based on your promise, and bad things can happen if you then break your promise.

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