Why can't you use scientific notation in kernel

I was trying to write a kernel (4.8.1) module, and if I use

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));
}

it raises this error

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

while if I use

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));
}

it works without a problem.

So why can't you use scientific notation in the kernel? I mean, I think is much easier and more readable to use 1e3; 1e6; 1e9 instead of 1000; 1000000; 1000000000.

Is just a matter of portability/robustness?
Or something like (in this case)

You need ns? Use ktime_to_ns
You need us? Use ktime_to_us
You need ms? Use ktime_to_ms

P.S. I tried with a simple .c program and it works without problems

#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;
}

Solution

1e3 is not equivalent to 1000.

1000 is an integer constant of type int. 1e3 is a floating-point constant of type double, equivalent to 1000.0. That makes remaining/1e3 a floating-point division, which is what the compiler is complaining about.