• ia64
• SystemCalls

IA-64 : System Calls

I (=Arun) was examining the consequences of using ar.k5 as the system call entry point.

Key findings:

1. Latency of reading ar.k5 is 12 cycles as opposed to 1 cycles for other GRs
2. Using an immediate value or loading it from memory saves about 11 cycles.

test.c

<pre> #include <stdio.h>

#define COUNT 10000 typedef unsigned long uint64_t;

static inline uint64_t ia64_get_itc() {

• uint64_t res;

  asm volatile ("mov %0=ar.itc" : "=r" (res)); return res;

}

static inline uint64_t ia64_get_k5() {

• uint64_t res;

  asm volatile ("mov %0=ar.k5" : "=r" (res)); return res;

}

static inline uint64_t ia64_get_r1() {

• uint64_t res;

  asm volatile ("mov %0=r1" : "=r" (res)); return res;

}

main() {

• uint64_t t1, t2, i; t1 = ia64_get_itc();

  for (i = 0; i < COUNT; i++)

  • ia64_get_k5();
  t2 = ia64_get_itc(); printf("%d\n", (t2 - t1)/ COUNT); t1 = ia64_get_itc();

  for (i = 0; i < COUNT; i++)

  • ia64_get_r1();
  t2 = ia64_get_itc(); printf("%d\n", (t2 - t1)/ COUNT); t1 = ia64_get_itc();

  for (i = 0; i < COUNT; i++)

  • getpid1();
  t2 = ia64_get_itc(); printf("%d\n", (t2 - t1)/ COUNT); t1 = ia64_get_itc();

  for (i = 0; i < COUNT; i++)

  • getpid2();
  t2 = ia64_get_itc(); printf("%d\n", (t2 - t1)/ COUNT); t1 = ia64_get_itc();

  for (i = 0; i < COUNT; i++)

  • getpid3();
  t2 = ia64_get_itc(); printf("%d\n", (t2 - t1)/ COUNT);

} </pre>

t.S

<pre> #include <machine/asm.h>

entry: data8 0xa000000000000000

.text

_error:

• mov r14=ip ;; mov b0=r14 br.cond.sptk.few b0

ENTRY(getpid1, 8)

• alloc r9=ar.pfs,8,0,0,0 mov.m r31=ar.k5 mov r10=b0;; mov r8=20 mov b7=r31 br.call.sptk.few b0=b7;; cmp.eq p0,p6=r0,r10

(p6) br.cond.sptk.few _error;

• br.ret.sptk.few b0;;

.endp

ENTRY(getpid2, 8)

• alloc r9=ar.pfs,8,0,0,0 movl r31=0xa000000000000000 mov r10=b0;; mov r8=20 mov b7=r31 br.call.sptk.few b0=b7;; cmp.eq p0,p6=r0,r10

(p6) br.cond.sptk.few _error

• br.ret.sptk.few b0;;

.endp

ENTRY(getpid3, 8)

• alloc r9=ar.pfs,8,0,0,0 addl r14 = @gprel(entry#), gp;; ld8 r31 = [r14] mov r10=b0;; mov r8=20 mov b7=r31 br.call.sptk.few b0=b7;; cmp.eq p0,p6=r0,r10

(p6) br.cond.sptk.few _error

• br.ret.sptk.few b0;;

.endp </pre>

Results

<pre> $ gcc -O2 test.c t.S -o t $ ./t 12 1 433 422 422 </pre>