Exactly. The irony in all of this is that C is not a portable assembler. If you want the exact semantics of a hardware instruction, you cannot get it, because the compiler reasons with C's abstract machine that assumes your program doesn't have undefined behavior, like signed wraparound, when in some situations you in fact do want signed wraparound, since that's what literally every modern CPU does.

I think that the undefinde behaviour should be partially specified. In the case you describe, it should require that it must do one of the following:

1. Return any 32-bit answer for the right shift. (The final result will be zero due to the bitwise AND, though, regardless of the intermediate answer.) The intermediate answer must be "frozen" so that if it is assigned to a variable and then used multiple times without writing to that variable again then you will get the same answer each time.

2. Result in a run-time error when that code is reached.

3. Result in a compile-time error (only valid if the compiler can determine for sure that the program would run with a shift amount out of range, e.g. if the shift amount is a constant).

4. Have a behaviour which depends on the underlying instruction set (whatever the right shift instruction does in that instruction set when given a shift amount which is out of range), if it is defined. (A compiler switch may be provided to switch between this and other behaviours.) In this case, if optimization is enabled then there may be some strange cases with some instruction sets where the optimizer makes an assumption which is not valid, but bad assumptions such as this should be reduced if possible and reasonable to do so.

In all cases, a compiler warning may be given (if enabled and detected by the compiler), in addition to the effects above.

I wanted to reply that your point 3 should already be possible with C++ constexpr functions because it doesn't allow undefined behavior. But I it seems I was wrong about that or maybe I'm doing it wrong:

    [[nodiscard]] constexpr uint64_t
    getBits( uint8_t nBits )
    {
        return BITBUFFER >> ( 64 - nBits ) & ( ( 1ULL << nBits ) - 1U );
    }
 
    int main()
    {
        std::cerr << getBits( 0 ) << "\n";
        std::cerr << getBits( 1 ) << "\n";
        return 0;
    }

The first output will print a random number, 140728069214376 in my case, while the second line will always print 1. However, when I put the ( ( 1ULL << nBits ) - 1U ) part into a separate function and print the values for that, then getBits( 0 ) suddenly always returns 0 as if the compiler understands suddenly that it will and with 0.

    template<uint8_t nBits>
    [[nodiscard]] constexpr uint64_t
    getBits2()
    {
        return BITBUFFER >> ( 64 - nBits ) & ( ( 1ULL << nBits ) - 1U );
    }

In this case, the compiler will only print a warning when trying to call it with getBits2<0>. And here I kinda thought that constexpr would lead to errors on undefined behavior, partly because it always complains about uninitialized std::array local variables being an error. That seems inconsistent to me. Well, I guess that's what -Werror is for ...

Compiled with -std=c++17 and clang 16.0.0 on godbolt: https://godbolt.org/z/qxxWW93Tx

Unfortunately constexpr doesn't imply constant evaluation. Your function can still potentially be executed at runtime.

If you use the result in an expression that requires a constant (an array bound, a non-type template parameter, a static_assert, or, in c++20, to initialize a constinit variable), then that will force constant evaluation and you'll see the error.

Having said that, compilers have bugs (or simply not fully implemented features), so it is certainly possible that both GCC and clang will fail to correctly catch constant time evaluation UB in some circumstances.

Ah thanks, I was not aware that these compile-time checks are only done when it is evaluated in a compile-time evaluating context.

To add to your list, using C++20 consteval instead of constexpr also triggers the error.