Does “default” switch case disturb jump table optimization?

问题

In my code I'm used to write fall-back default cases containing asserts like the following, to guard me against forgetting to update the switch in case semantics change

switch(mode) {
case ModeA: ... ;
case ModeB: ... ;
case .. /* many of them ... */
default: {
  assert(0 && "Unknown mode!");
  return ADummyValue();
}
};

Now I wonder whether the artificial fall-back check default case will interfere with jump table generations? Imagine "ModeA" an "ModeB" etc are consecutive so the compiler could optimize into a table. Since the "default" case contains an actual "return" statement (since the assert will disappear in release mode and the compiler will moan about a missing return statement), it seems unlikely the compiler optimizes the default branch away.

What's the best way to handle this? Some friend recommended me to replace "ADummyValue" with a null pointer dereference, so that the compiler, in presence of undefined behavior, could omit to warn about a missing return statement. Are there better ways to solve this?

回答1:

If your compiler is MSVC, you can use __assume intrinsic : http://msdn.microsoft.com/en-us/library/1b3fsfxw(v=VS.80).aspx

回答2:

At least with the compilers I've looked at, the answer is generally no. Most of them will compile a switch statement like this to code roughly equivalent to:

if (mode < modeA || mode > modeLast) {
    assert(0 && "Unknown mode!");
    return ADummyValue();
}
switch(mode) { 
    case modeA: ...;
    case modeB: ...;
    case modeC: ...;
    // ...
    case modeLast: ...;
}

回答3:

if you're using "default" (ha!) <assert.h>, the definition's tied to the NDEBUG macro anyway, so maybe just

    case nevermind:
#if !defined(NDEBUG)
    default:
        assert("can" && !"happen");
#endif
    }

回答4:

I only see 1 solution in case the optimization actually is disturbed: the infamous "#ifndef NDEBUG" round the default case. Not the nicest trick, but clear in this situation.

BTW: did you already have a look what your compiler does with and without the default case?

回答5:

If you have a state that should never be reached, then you should kill the program, because it just reached an unexpected state, even in the release mode (you might just be more diplomatic and actually save users data and do all that other nice stuff before going down).

And please don't obsess over micro optimizations unless you actually have measured (using a profiler) that you need them.

回答6:

The best way to handle this is not to disable the assert. That way you can also keep an eye on possible bugs. Sometimes it is better for the application to crash with a good message explaining what exactly happened, then to continue working.

回答7:

Use compiler extensions:

// assume.hpp
#pragma once

#if defined _MSC_VER
#define MY_ASSUME(e) (__assume(e), (e) ? void() : void())
#elif defined __GNUC__
#define MY_ASSUME(e) ((e) ? void() : __builtin_unreachable())
#else   // defined __GNUC__
#error unknown compiler
#endif  // defined __GNUC__

// assert.hpp
#include <cassert>
#include "assume.hpp"

#undef MY_ASSERT
#ifdef NDEBUG
#define MY_ASSERT MY_ASSUME
#else   // NDEBUG
#define MY_ASSERT assert
#endif  // NDEBUG

来源：https://stackoverflow.com/questions/4278807/does-default-switch-case-disturb-jump-table-optimization

标签

c++

switch-statement

jump-table