为什么你不能在内核中使用科学记数法

Question

我正在尝试编写内核(4.8.1)模块,如果我使用的话

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));
}

它引发了这个错误

error: SSE register return with SSE disabled
   printk("\t\t(%llu ns; %llu us)\n",
   ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
          remaining,
          ~~~~~~~~~~
          (unsigned long long) (remaining/1e3));
          ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

如果我使用

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));
}

它没有问题.

那你为什么不能在内核中使用科学记数法呢？我的意思是,我认为是更容易和更可读的使用1e3; 1e6; 1e9,而不是1000; 1000000; 1000000000.

只是可移植性/健壮性问题？
或类似的东西(在这种情况下)

你需要ns吗？使用ktime_to_ns
你需要我们吗？使用ktime_to_us
你需要ms？使用ktime_to_ms

PS我试过一个简单的.c程序,它没有问题

#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;
}

Answer 1

1e3是不是等同于1000.

1000是类型的整数常量int.1e3是一个类型的浮点常量double,相当于1000.0.这是remaining/1e3一个浮点除法,这是编译器所抱怨的.

另请参见禁用SSE的SSE寄存器返回.