4
我想寫一個內核(4.8.1)模塊,並且如果我用爲什麼你不能用科學記數法在內核
if (hrtimer_cancel(&hr_timer) == 1) {
u64 remaining = ktime_to_ns(hrtimer_get_remaining(&hr_timer));
printk("(%llu ns; %llu us)\n", remaining,
(unsigned long long) (remaining/1e3));
}
它提出了這個錯誤
error: SSE register return with SSE disabled
printk("\t\t(%llu ns; %llu us)\n",
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
remaining,
~~~~~~~~~~
(unsigned long long) (remaining/1e3));
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
而如果我用
if (hrtimer_cancel(&hr_timer) == 1) {
u64 remaining = ktime_to_ns(hrtimer_get_remaining(&hr_timer));
printk("(%llu ns; %llu us)\n", remaining,
(unsigned long long) (remaining/1000));
}
它工作沒有問題。
那麼爲什麼你不能在內核中使用科學記數法?我的意思是,我認爲使用1e3; 1e6; 1e9
而不是1000; 1000000; 1000000000
更容易和更具可讀性。
只是一個可移植性/健壯性的問題?
或類似的東西(在這種情況下)
你需要ns?使用
ktime_to_ns
你需要我們嗎?使用ktime_to_us
你需要ms?使用ktime_to_ms
P.S.我試圖用一個簡單的.C程序,它的工作原理沒有問題
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
void error_handler(const char *msg)
{
perror(msg);
exit(EXIT_FAILURE);
}
unsigned long parse_num(const char *number)
{
unsigned long v;
char *p;
errno = 0;
v = strtoul(number, &p, 10);
if (errno != 0 || *p != '\0')
error_handler("parse_num | strtoul");
return v;
}
int main(int argc, char *argv[])
{
if (argc != 2)
{
fprintf(stderr, "Usage: %s number_greater_than_1000\n", argv[0]);
return EXIT_FAILURE;
}
unsigned long number = parse_num(argv[1]);
if (number < 1e3 || number > 1e6)
{
fprintf(stderr, "Need to be a number in range (%lu, %lu)\n", (unsigned long) 1e3, (unsigned long) 1e6);
return EXIT_FAILURE;
}
printf("Original: %lu\tScaled: %lu\n", number, (unsigned long) (number/1e3));
return EXIT_SUCCESS;
}
我認爲主要問題不是符號,而是'1e3'是一個'double'文字。內核通常不支持浮點以節省省時寄存器。 – Banex
[SSE禁用SSE寄存器返回]的可能重複(http://stackoverflow.com/questions/1556142/sse-register-return-with-sse-disabled) – 2016-11-20 18:22:07