2016-11-20 21 views
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; 
} 
+6

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

+1

[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

回答

6

1e3相當於1000

1000是類型爲int的整數常量。 1e3double類型的浮點常量,相當於1000.0。這使得remaining/1e3成爲浮點部分,這正是編譯器所抱怨的。請參閱SSE register return with SSE disabled