问题
This is a strange request but I have a feeling that it could be possible. What I would like is to insert some pragmas or directives into areas of my code (written in C) so that GCC's register allocator will not use them.
I understand that I can do something like this, which might set aside this register for this variable
register int var1 asm ("EBX") = 1984;
register int var2 asm ("r9") = 101;
The problem is that I'm inserting new instructions (for a hardware simulator) directly and GCC and GAS don't recognise these yet. My new instructions can use the existing general purpose registers and I want to make sure that I have some of them (i.e. r12->r15) reserved.
Right now, I'm working in a mockup environment and I want to do my experiments quickly. In the future I will append GAS and add intrinsics into GCC, but right now I'm looking for a quick fix.
Thanks!
回答1:
When writing GCC inline assembler, you can specify a "clobber list" - a list of registers that may be overwritten by your inline assembler code. GCC will then do whatever is needed to save and restore data in those registers (or avoid their use in the first place) over the course of the inline asm segment. You can also bind input or output registers to C variables.
For example:
inline unsigned long addone(unsigned long v)
{
unsigned long rv;
asm("mov $1, %%eax;"
"mov %0, %%ebx;"
"add %%eax, %%ebx"
: /* outputs */ "b" (rv)
: /* inputs */ "g" (v) /* select unused general purpose reg into %0 */
: /* clobbers */ "eax"
);
}
For more information, see the GCC-Inline-Asm-HOWTO.
回答2:
If you use global explicit register variables, these will be reserved throughout the compilation unit, and will not be used by the compiler for anything else (it may still be used by the system's libraries, so choose something that will be restored by those). local register variables do not guarantee that your value will be in the register at all times, but only when referenced by code or as an asm operand.
回答3:
If you write an inline asm block for your new instructions, there are commands that inform GCC what registers are used by that block and how they are used. GCC will then avoid using those registers or will at least save and reload their contents.
回答4:
Non-hardcoded scratch register in inline assembly
This is not a direct answer to the original question, but since and since I keep Googling this in that context and since https://stackoverflow.com/a/6683183/895245 was accepted, I'm going to try and provide a possible improvement to that answer.
The improvement is the following: you should avoid hard-coding your scratch registers when possible, to give the register allocator more freedom.
Therefore, as an educational example that is useless in practice (could be done in a single lea (%[in1], %[in2]), %[out];), the following hardcoded scratch register code:
bad.c
#include <assert.h>
#include <inttypes.h>
int main(void) {
uint64_t in1 = 0xFFFFFFFF;
uint64_t in2 = 1;
uint64_t out;
__asm__ (
"mov %[in2], %%rax;" /* scratch = in2 */
"add %[in1], %%rax;" /* scratch += in1 */
"mov %%rax, %[out];" /* out = scratch */
: [out] "=r" (out)
: [in1] "r" (in1),
[in2] "r" (in2)
: "rax"
);
assert(out == 0x100000000);
}
could compile to something more efficient if you instead use this non-hardcoded version:
good.c
#include <assert.h>
#include <inttypes.h>
int main(void) {
uint64_t in1 = 0xFFFFFFFF;
uint64_t in2 = 1;
uint64_t out;
uint64_t scratch;
__asm__ (
"mov %[in2], %[scratch];" /* scratch = in2 */
"add %[in1], %[scratch];" /* scratch += in1 */
"mov %[scratch], %[out];" /* out = scratch */
: [scratch] "=&r" (scratch),
[out] "=r" (out)
: [in1] "r" (in1),
[in2] "r" (in2)
:
);
assert(out == 0x100000000);
}
since the compiler is free to choose any register it wants instead of just rax,
Note that in this example we had to mark the scratch as an early clobber register with & to prevent it from being put into the same register as an input, I have explained that in more detail at: When to use earlyclobber constraint in extended GCC inline assembly? This example also happens to fail in the implementation I tested on without &.
Tested in Ubuntu 18.10 amd64, GCC 8.2.0, compile and run with:
gcc -O3 -std=c99 -ggdb3 -Wall -Werror -pedantic -o good.out good.c
./good.out
Non-hardcoded scratch registers are also mentioned in the GCC manual 6.45.2.6 "Clobbers and Scratch Registers", although their example is too much for mere mortals to take in at once:
Rather than allocating fixed registers via clobbers to provide scratch registers for an asm statement, an alternative is to define a variable and make it an early-clobber output as with a2 and a3 in the example below. This gives the compiler register allocator more freedom. You can also define a variable and make it an output tied to an input as with a0 and a1, tied respectively to ap and lda. Of course, with tied outputs your asm can’t use the input value after modifying the output register since they are one and the same register. What’s more, if you omit the early-clobber on the output, it is possible that GCC might allocate the same register to another of the inputs if GCC could prove they had the same value on entry to the asm. This is why a1 has an early-clobber. Its tied input, lda might conceivably be known to have the value 16 and without an early-clobber share the same register as %11. On the other hand, ap can’t be the same as any of the other inputs, so an early-clobber on a0 is not needed. It is also not desirable in this case. An early-clobber on a0 would cause GCC to allocate a separate register for the "m" ((const double ()[]) ap) input. Note that tying an input to an output is the way to set up an initialized temporary register modified by an asm statement. An input not tied to an output is assumed by GCC to be unchanged, for example "b" (16) below sets up %11 to 16, and GCC might use that register in following code if the value 16 happened to be needed. You can even use a normal asm output for a scratch if all inputs that might share the same register are consumed before the scratch is used. The VSX registers clobbered by the asm statement could have used this technique except for GCC’s limit on the number of asm parameters.
static void dgemv_kernel_4x4 (long n, const double *ap, long lda, const double *x, double *y, double alpha) { double *a0; double *a1; double *a2; double *a3; __asm__ ( /* lots of asm here */ "#n=%1 ap=%8=%12 lda=%13 x=%7=%10 y=%0=%2 alpha=%9 o16=%11\n" "#a0=%3 a1=%4 a2=%5 a3=%6" : "+m" (*(double (*)[n]) y), "+&r" (n), // 1 "+b" (y), // 2 "=b" (a0), // 3 "=&b" (a1), // 4 "=&b" (a2), // 5 "=&b" (a3) // 6 : "m" (*(const double (*)[n]) x), "m" (*(const double (*)[]) ap), "d" (alpha), // 9 "r" (x), // 10 "b" (16), // 11 "3" (ap), // 12 "4" (lda) // 13 : "cr0", "vs32","vs33","vs34","vs35","vs36","vs37", "vs40","vs41","vs42","vs43","vs44","vs45","vs46","vs47" ); }
来源：https://stackoverflow.com/questions/6682733/gcc-prohibit-use-of-some-registers