在Sparc 32位上处理值> 2 ^ 32的整数
you*_*t13 12 c assembly sparc long-integer
我编写了一个小程序来测量循环所花费的时间(通过内联Sparc汇编代码片段).
一切都是正确的,直到我将迭代次数设置在大约4.0 + 9(高于2 ^ 32)之上.
这是代码片段:
#include <stdio.h>
#include <sys/time.h>
#include <unistd.h>
#include <math.h>
#include <stdint.h>
int main (int argc, char *argv[])
{
// For indices
int i;
// Set the number of executions
int nRunning = atoi(argv[1]);
// Set the sums
double avgSum = 0.0;
double stdSum = 0.0;
// Average of execution time
double averageRuntime = 0.0;
// Standard deviation of execution time
double deviationRuntime = 0.0;
// Init sum
unsigned long long int sum = 0;
// Number of iterations
unsigned long long int nLoop = 4000000000ULL;
//uint64_t nLoop = 4000000000;
// DEBUG
printf("sizeof(unsigned long long int) = %zu\n",sizeof(unsigned long long int));
printf("sizeof(unsigned long int) = %zu\n",sizeof(unsigned long int));
// Time intervals
struct timeval tv1, tv2;
double diff;
// Loop for multiple executions
for (i=0; i<nRunning; i++)
{
// Start time
gettimeofday (&tv1, NULL);
// Loop with Sparc assembly into C source
asm volatile ("clr %%g1\n\t"
"clr %%g2\n\t"
"mov %1, %%g1\n" // %1 = input parameter
"loop:\n\t"
"add %%g2, 1, %%g2\n\t"
"subcc %%g1, 1, %%g1\n\t"
"bne loop\n\t"
"nop\n\t"
"mov %%g2, %0\n" // %0 = output parameter
: "=r" (sum) // output
: "r" (nLoop) // input
: "g1", "g2"); // clobbers
// End time
gettimeofday (&tv2, NULL);
// Compute runtime for loop
diff = (tv2.tv_sec - tv1.tv_sec) * 1000000ULL + (tv2.tv_usec - tv1.tv_usec);
// Summing diff time
avgSum += diff;
stdSum += (diff*diff);
// DEBUG
printf("diff = %e\n", diff);
printf("avgSum = %e\n", avgSum);
}
// Compute final averageRuntime
averageRuntime = avgSum/nRunning;
// Compute standard deviation
deviationRuntime = sqrt(stdSum/nRunning-averageRuntime*averageRuntime);
// Print results
printf("(Average Elapsed time, Standard deviation) = %e usec %e usec\n", averageRuntime, deviationRuntime);
// Print sum from assembly loop
printf("Sum = %llu\n", sum);
例如,使用nLoop<2 ^ 32,我得到正确的值diff,avgSum和stdSum.确实,printf有nLoop = 4.0e+9,给,给:
sizeof(unsigned long long int) = 8
sizeof(unsigned long int) = 4
diff = 9.617167e+06
avgSum = 9.617167e+06
diff = 9.499878e+06
avgSum = 1.911704e+07
(Average Elapsed time, Standard deviation) = 9.558522e+06 usec 5.864450e+04 usec
Sum = 4000000000
该代码被编译在Debian Sparc 32 bits Etch用gcc 4.1.2.
不幸的是,如果我举个例子nLoop = 5.0e+9,我会得到小而不正确的测量时间值; 这是这种情况下的printf输出:
sizeof(unsigned long long int) = 8
sizeof(unsigned long int) = 4
diff = 5.800000e+01
avgSum = 5.800000e+01
diff = 4.000000e+00
avgSum = 6.200000e+01
(Average Elapsed time, Standard deviation) = 3.100000e+01 usec 2.700000e+01 usec
Sum = 5000000000
我不知道问题可能来自哪里,我已经做了其他测试,uint64_t但没有成功.
也许问题是我处理large integers (> 2^32)32位操作系统或者它可能是汇编内联代码,它不支持8字节整数.
如果有人能给我一些线索来解决这个错误,
问候
更新1:
按照建议@Andrew Henle,我采用相同的代码,但不是内联Sparc Assembly片段,我只是简单地循环.
这是带有简单循环的程序nLoop = 5.0e+9(参见" unsigned long long int nLoop = 5000000000ULL;" 行,所以上面是limit 2^32-1:
#include <stdio.h>
#include <stdlib.h>
#include <sys/time.h>
#include <unistd.h>
#include <math.h>
#include <stdint.h>
int main (int argc, char *argv[])
{
// For indices of nRunning
int i;
// For indices of nRunning
unsigned long long int j;
// Set the number of executions
int nRunning = atoi(argv[1]);
// Set the sums
unsigned long long int avgSum = 0;
unsigned long long int stdSum = 0;
// Average of execution time
double averageRuntime = 0.0;
// Standard deviation of execution time
double deviationRuntime = 0.0;
// Init sum
unsigned long long int sum;
// Number of iterations
unsigned long long int nLoop = 5000000000ULL;
// DEBUG
printf("sizeof(unsigned long long int) = %zu\n",sizeof(unsigned long long int));
printf("sizeof(unsigned long int) = %zu\n",sizeof(unsigned long int));
// Time intervals
struct timeval tv1, tv2;
unsigned long long int diff;
// Loop for multiple executions
for (i=0; i<nRunning; i++)
{
// Reset sum
sum = 0;
// Start time
gettimeofday (&tv1, NULL);
// Loop with Sparc assembly into C source
/* asm volatile ("clr %%g1\n\t"
"clr %%g2\n\t"
"mov %1, %%g1\n" // %1 = input parameter
"loop:\n\t"
"add %%g2, 1, %%g2\n\t"
"subcc %%g1, 1, %%g1\n\t"
"bne loop\n\t"
"nop\n\t"
"mov %%g2, %0\n" // %0 = output parameter
: "=r" (sum) // output
: "r" (nLoop) // input
: "g1", "g2"); // clobbers
*/
// Classic loop
for (j=0; j<nLoop; j++)
sum ++;
// End time
gettimeofday (&tv2, NULL);
// Compute runtime for loop
diff = (unsigned long long int) ((tv2.tv_sec - tv1.tv_sec) * 1000000 + (tv2.tv_usec - tv1.tv_usec));
// Summing diff time
avgSum += diff;
stdSum += (diff*diff);
// DEBUG
printf("diff = %llu\n", diff);
printf("avgSum = %llu\n", avgSum);
printf("stdSum = %llu\n", stdSum);
// Print sum from assembly loop
printf("Sum = %llu\n", sum);
}
// Compute final averageRuntime
averageRuntime = avgSum/nRunning;
// Compute standard deviation
deviationRuntime = sqrt(stdSum/nRunning-averageRuntime*averageRuntime);
// Print results
printf("(Average Elapsed time, Standard deviation) = %e usec %e usec\n", averageRuntime, deviationRuntime);
return 0;
}
此代码段工作正常,即变量sum打印为(请参阅" printf("Sum = %llu\n", sum)";):
Sum = 5000000000
所以问题来自Sparc Assembly block的版本.
我怀疑,在该汇编代码,行"mov %1, %%g1\n" // %1 = input parameter要差些存储nLoop成%g1 register(我认为这%g1是一个32位寄存器,因此不能存储上述值2^32-1).
但是,行的ouput参数(变量sum):
"mov %%g2, %0\n" // %0 = output parameter
超过限制,因为它等于5000000000.
我在带有Assembly循环的版本之间附加了vimdiff,没有它:
在左边,使用Assembly编程,在右边,没有Assembly(只是一个简单的循环)
我提醒你我的问题是,对于nLoop> 2 ^ 32-1并且使用汇编循环,我sum在执行结束时得到一个有效参数但是没有效(太短)average和standard deviation时间(花在循环中); 这是一个输出示例nLoop = 5000000000ULL:
sizeof(unsigned long long int) = 8
sizeof(unsigned long int) = 4
diff = 17
avgSum = 17
stdSum = 289
Sum = 5000000000
diff = 4
avgSum = 21
stdSum = 305
Sum = 5000000000
(Average Elapsed time, Standard deviation) = 1.000000e+01 usec 7.211103e+00 usec
通过拍摄nLoop = 4.0e+9,即nLoop = 4000000000ULL没有问题,时间值是有效的.
更新2:
我正在通过生成汇编代码进行更深入的搜索.版本nLoop = 4000000000 (4.0e+9)如下:
.file "loop-WITH-asm-inline-4-Billions.c"
.section ".rodata"
.align 8
.LLC1:
.asciz "sizeof(unsigned long long int) = %zu\n"
.align 8
.LLC2:
.asciz "sizeof(unsigned long int) = %zu\n"
.align 8
.LLC3:
.asciz "diff = %llu\n"
.align 8
.LLC4:
.asciz "avgSum = %llu\n"
.align 8
.LLC5:
.asciz "stdSum = %llu\n"
.align 8
.LLC6:
.asciz "Sum = %llu\n"
.global __udivdi3
.global __cmpdi2
.global __floatdidf
.align 8
.LLC7:
.asciz "(Average Elapsed time, Standard deviation) = %e usec %e usec\n"
.align 8
.LLC0:
.long 0
.long 0
.section ".text"
.align 4
.global main
.type main, #function
.proc 04
main:
save %sp, -248, %sp
st %i0, [%fp+68]
st %i1, [%fp+72]
ld [%fp+72], %g1
add %g1, 4, %g1
ld [%g1], %g1
mov %g1, %o0
call atoi, 0
nop
mov %o0, %g1
st %g1, [%fp-68]
st %g0, [%fp-64]
st %g0, [%fp-60]
st %g0, [%fp-56]
st %g0, [%fp-52]
sethi %hi(.LLC0), %g1
or %g1, %lo(.LLC0), %g1
ldd [%g1], %f8
std %f8, [%fp-48]
sethi %hi(.LLC0), %g1
or %g1, %lo(.LLC0), %g1
ldd [%g1], %f8
std %f8, [%fp-40]
mov 0, %g2
sethi %hi(4000000000), %g3
std %g2, [%fp-24]
sethi %hi(.LLC1), %g1
or %g1, %lo(.LLC1), %o0
mov 8, %o1
call printf, 0
nop
sethi %hi(.LLC2), %g1
or %g1, %lo(.LLC2), %o0
mov 4, %o1
call printf, 0
nop
st %g0, [%fp-84]
b .LL2
nop
.LL3:
st %g0, [%fp-32]
st %g0, [%fp-28]
add %fp, -92, %g1
mov %g1, %o0
mov 0, %o1
call gettimeofday, 0
nop
ldd [%fp-24], %o4
clr %g1
clr %g2
mov %o4, %g1
loop:
add %g2, 1, %g2
subcc %g1, 1, %g1
bne loop
nop
mov %g2, %o4
std %o4, [%fp-32]
add %fp, -100, %g1
mov %g1, %o0
mov 0, %o1
call gettimeofday, 0
nop
ld [%fp-100], %g2
ld [%fp-92], %g1
sub %g2, %g1, %g2
sethi %hi(999424), %g1
or %g1, 576, %g1
smul %g2, %g1, %g3
ld [%fp-96], %g2
ld [%fp-88], %g1
sub %g2, %g1, %g1
add %g3, %g1, %g1
st %g1, [%fp-12]
sra %g1, 31, %g1
st %g1, [%fp-16]
ldd [%fp-64], %o4
ldd [%fp-16], %g2
addcc %o5, %g3, %g3
addx %o4, %g2, %g2
std %g2, [%fp-64]
ld [%fp-16], %g2
ld [%fp-12], %g1
smul %g2, %g1, %g4
ld [%fp-16], %g2
ld [%fp-12], %g1
smul %g2, %g1, %g1
add %g4, %g1, %g4
ld [%fp-12], %g2
ld [%fp-12], %g1
umul %g2, %g1, %g3
rd %y, %g2
add %g4, %g2, %g4
mov %g4, %g2
ldd [%fp-56], %o4
addcc %o5, %g3, %g3
addx %o4, %g2, %g2
std %g2, [%fp-56]
sethi %hi(.LLC3), %g1
or %g1, %lo(.LLC3), %o0
ld [%fp-16], %o1
ld [%fp-12], %o2
call printf, 0
nop
sethi %hi(.LLC4), %g1
or %g1, %lo(.LLC4), %o0
ld [%fp-64], %o1
ld [%fp-60], %o2
call printf, 0
nop
sethi %hi(.LLC5), %g1
or %g1, %lo(.LLC5), %o0
ld [%fp-56], %o1
ld [%fp-52], %o2
call printf, 0
nop
sethi %hi(.LLC6), %g1
or %g1, %lo(.LLC6), %o0
ld [%fp-32], %o1
ld [%fp-28], %o2
call printf, 0
nop
ld [%fp-84], %g1
add %g1, 1, %g1
st %g1, [%fp-84]
.LL2:
ld [%fp-84], %g2
ld [%fp-68], %g1
cmp %g2, %g1
bl .LL3
nop
ld [%fp-68], %g1
sra %g1, 31, %g1
ld [%fp-68], %g3
mov %g1, %g2
ldd [%fp-64], %o0
mov %g2, %o2
mov %g3, %o3
call __udivdi3, 0
nop
mov %o0, %g2
mov %o1, %g3
std %g2, [%fp-136]
ldd [%fp-136], %o0
mov 0, %o2
mov 0, %o3
call __cmpdi2, 0
nop
mov %o0, %g1
cmp %g1, 1
bl .LL6
nop
ldd [%fp-136], %o0
call __floatdidf, 0
nop
std %f0, [%fp-144]
b .LL5
nop
.LL6:
ldd [%fp-136], %o4
and %o4, 0, %g2
and %o5, 1, %g3
ld [%fp-136], %o5
sll %o5, 31, %g1
ld [%fp-132], %g4
srl %g4, 1, %o5
or %o5, %g1, %o5
ld [%fp-136], %g1
srl %g1, 1, %o4
or %g2, %o4, %g2
or %g3, %o5, %g3
mov %g2, %o0
mov %g3, %o1
call __floatdidf, 0
nop
std %f0, [%fp-144]
ldd [%fp-144], %f8
ldd [%fp-144], %f10
faddd %f8, %f10, %f8
std %f8, [%fp-144]
.LL5:
ldd [%fp-144], %f8
std %f8, [%fp-48]
ld [%fp-68], %g1
sra %g1, 31, %g1
ld [%fp-68], %g3
mov %g1, %g2
ldd [%fp-56], %o0
mov %g2, %o2
mov %g3, %o3
call __udivdi3, 0
nop
mov %o0, %g2
mov %o1, %g3
std %g2, [%fp-128]
ldd [%fp-128], %o0
mov 0, %o2
mov 0, %o3
call __cmpdi2, 0
nop
mov %o0, %g1
cmp %g1, 1
bl .LL8
nop
ldd [%fp-128], %o0
call __floatdidf, 0
nop
std %f0, [%fp-120]
b .LL7
nop
.LL8:
ldd [%fp-128], %o4
and %o4, 0, %g2
and %o5, 1, %g3
ld [%fp-128], %o5
sll %o5, 31, %g1
ld [%fp-124], %g4
srl %g4, 1, %o5
or %o5, %g1, %o5
ld [%fp-128], %g1
srl %g1, 1, %o4
or %g2, %o4, %g2
or %g3, %o5, %g3
mov %g2, %o0
mov %g3, %o1
call __floatdidf, 0
nop
std %f0, [%fp-120]
ldd [%fp-120], %f8
ldd [%fp-120], %f10
faddd %f8, %f10, %f8
std %f8, [%fp-120]
.LL7:
ldd [%fp-48], %f8
ldd [%fp-48], %f10
fmuld %f8, %f10, %f8
ldd [%fp-120], %f10
fsubd %f10, %f8, %f8
std %f8, [%fp-112]
ldd [%fp-112], %f8
fsqrtd %f8, %f8
std %f8, [%fp-152]
ldd [%fp-152], %f10
ldd [%fp-152], %f8
fcmpd %f10, %f8
nop
fbe .LL9
nop
ldd [%fp-112], %o0
call sqrt, 0
nop
std %f0, [%fp-152]
.LL9:
ldd [%fp-152], %f8
std %f8, [%fp-40]
sethi %hi(.LLC7), %g1
or %g1, %lo(.LLC7), %o0
ld [%fp-48], %o1
ld [%fp-44], %o2
ld [%fp-40], %o3
ld [%fp-36], %o4
call printf, 0
nop
mov 0, %g1
mov %g1, %i0
restore
jmp %o7+8
nop
.size main, .-main
.ident "GCC: (GNU) 4.1.2 20061115 (prerelease) (Debian 4.1.1-21)"
.section ".note.GNU-stack"
当我生成汇编代码版本时nLoop = 5000000000 (5.0e+9),差异如下图所示(带vimdiff):
"4亿"版块:
mov 0, %g2
sethi %hi(4000000000), %g3
被"5亿"版本替换为:
mov 1, %g2
sethi %hi(705032192), %g3
or %g3, 512, %g3
我可以看到5.0+e9,由于指令,不能在32位上编码
sethi %hi(705032192), %g3
矛盾的是,当我编译版本"5亿"汇编代码时,ouput参数sum计算得很好,即等于5 Billions,我无法解释它.
欢迎任何帮助或评论,谢谢.
很大程度上取决于您使用的 sparc 版本和 ABI。如果您使用的是 sparc v8 或更早版本，则您拥有仅包含 32 位寄存器的 32 位模式。在这种情况下，当您尝试将 5000000000 加载到 32 位寄存器时，它会失败并加载 5000000000 mod 2 32（即 705032704）。这似乎正在发生。
另一方面，如果您有一个以 32 位模式运行的 64 位 sparc 处理器（通常称为 v8plus），那么您可以使用 64 位寄存器，因此这可以工作。
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