Avoid range check on switch/case statement in gcc?

Question

[Edit: It seems this is an issue on gcc versions prior to 4.4, I got confused because of a gcc bugzilla entry reporting it for 4.5 (latest). Sorry, I should've tested with more recent versions. Still, the problem is somewhat valid as most people don't run gcc 4.4+.]

Is it possible to tell the compiler the variable used in a switch fits within the provided case statements? In particular if it's a small range and there's a jump table generated.

extern int a;
main()
{
        switch (a & 0x7) {   // 0x7  == 111  values are 0-7
        case 0: f0(); break;
        case 1: f1(); break;
        case 2: f2(); break;
        case 3: f3(); break;
        case 4: f4(); break;
        case 5: f5(); break;
        case 6: f6(); break;
        case 7: f7(); break;
        }
}

I tried xor'ing to low bits (as the example), using enums, using gcc_unreachable() to no avail. The generated code always checks if the variable is inside the range, adding a pointless branch conditional and moving away the jump table calculation code.

Note: this is in the innermost loop of a decoder, performance matters significantly.

It seems I'm not the only one.

There is no way to tell gcc that the default branch is never taken, although it will omit the default branch if it can prove that the value is never out of range based on earlier conditional checks.

So, how do you help gcc prove the variable fits and there's no default branch in the example above? (Without adding a conditional branch, of course.)

EDIT1: This was on OS X 10.6 Snow Leopard with GCC 4.2 (default from Xcode.) It didn't happen with GCC 4.4/4.3 in linux (reported by Nathon and Jens Gustedt.)

EDIT2: The functions in the example are there for readability, think those are inlined or just statements. Making a function call on x86 is expensive.

Also the example, as mentioned in the note, belongs inside a loop on data (big data.)

The generated code with gcc 4.2/OS X is:

[...]
andl    $7, %eax
cmpl    $7, %eax
ja  L11
mov %eax, %eax
leaq    L20(%rip), %rdx
movslq  (%rdx,%rax,4),%rax
addq    %rdx, %rax
jmp *%rax
.align 2,0x90
L20:
.long   L12-L20
.long   L13-L20
.long   L14-L20
.long   L15-L20
.long   L16-L20
.long   L17-L20
.long   L18-L20
.long   L19-L20
L19:
[...]

The problem lies on cmp $7, %eax / ja L11.

EDIT3:

OK, I'm going with the ugly solution and adding a special case for gcc versions below 4.4 using a different version without a switch and using goto and gcc's &&label extensions.

static void *jtb[] = { &&c_1, &&c_2, &&c_3, &&c_4, &&c_5, &&c_6, &&c_7, &&c_8 };
[...]
goto *jtb[a & 0x7];
[...]
while(0) {
c_1:
// something
break;
c_2:
// something
break;
[...]
}

Note the array of labels is static so it's not computed every call.

Thanks everyone for the great help! And sorry to those with valid answers who didn't get the mark :(

Answer 1

Have you tried declaring the switch variable as a bitfield?

struct Container {
  uint16_t a:3;
  uint16_t unused:13;
};

struct Container cont;

cont.a = 5;  /* assign some value */
switch( cont.a ) {
...
}

Hope this works!

Answer 2

Perhaps just use a default label for the fist or last case?

Answer 3

Perhaps you could use an array of function pointers instead of a switch ?

#include <stdio.h>

typedef void (*func)(void);

static void f0(void) { printf("%s\n", __FUNCTION__); }
static void f1(void) { printf("%s\n", __FUNCTION__); }
static void f2(void) { printf("%s\n", __FUNCTION__); }
static void f3(void) { printf("%s\n", __FUNCTION__); }
static void f4(void) { printf("%s\n", __FUNCTION__); }
static void f5(void) { printf("%s\n", __FUNCTION__); }
static void f6(void) { printf("%s\n", __FUNCTION__); }
static void f7(void) { printf("%s\n", __FUNCTION__); }

int main(void)
{
    const func f[8] = { f0, f1, f2, f3, f4, f5, f6, f7 };
    int i;

    for (i = 0; i < 8; ++i)
    {
        f[i]();
    }
    return 0;
}

Answer 4

This question is certainly interesting from the standpoint of a missed compiler optimization that is seemingly obvious to us, and I did spend considerable time trying to come up with a straightforward solution, largely out of personal curiousity.

That said, I have to admit I am highly skeptical that this additional instruction will ever result in a measurable performance difference in practice, especially on a new mac. If you have any significant amount of data, you'll be I/O bound, and a single instruction will never be your bottleneck. If you have a tiny amount of data, then you'll need to perform a lot lot lot of calculations repeatedly before a single instruction will become a bottleneck.

Would you post some code to show that there really is a performance difference? Or describe the code and data your working with?

Answer 5

I didn't try, but I'm not sure that gcc_unreachable does the same thing as __builtin_unreachable. Googling the two, gcc_unreachable appears to be designed as a as an assertion tool for development of GCC itself, perhaps with a branch prediction hint included, whereas __builtin_unreachable makes the program instantly undefined — which sounds like deleting the basic block, which is what you want.

http://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html#index-g_t_005f_005fbuiltin_005funreachable-3075