Why does printf show negative values for unsigned int? [duplicate]

问题

Possible Duplicate:
Unsigned long with negative value

I have written one kernel module which interrupts any system call, prints its current user_id and input parameters passed to the system call function. Among them, one is sys_ioctl() as below:

asmlinkage long sys_ioctl(unsigned int fd, unsigned int cmd,unsigned long arg);

which means all input parameters are unsigned int numbers.

But when I print input parameters, I get the following output:

 fd=21, cmd=-1072143871 and arg=3202983648 
 fd=21, cmd=-1072143871 and arg=3202983648 
 fd=21, cmd=-1072143871 and arg=3202983648 
 ----------

Here is my function definition:

asmlinkage long our_sys_ioctl(unsigned int fd ,  unsigned int cmd , unsigned long arg)
{
    uid_t gtuid ;
    gtuid = getuid_call();
    printk ("our_sys_ioctl ---> uid = %d with fd=%i, cmd=%i and arg=%lu \n ", gtuid, fd, cmd, arg);
    return original_call_ioctl(fd , cmd , arg);
}

Any idea why cmd value is negative and what these values mean?

回答1:

When you use %i to print the cmd you are casting it to signed int as @Mario previously guessed. This is why it is negative.

You need to use %u to do the printing for it to remain an unsigned int

d or i - Signed decimal integer

u - Unsigned decimal integer

(from: http://www.cplusplus.com/reference/cstdio/printf/)

This will work as you would expect.

printk ("our_sys_ioctl ---> uid = %d with fd=%i, cmd=%u and arg=%lu \n ", gtuid, fd, cmd, arg);
                                                    ^^^^

回答2:

You print with %i which converts the argument to a signed int. Print with %u.

来源：https://stackoverflow.com/questions/14181691/why-does-printf-show-negative-values-for-unsigned-int

标签

Linux

linux-kernel

system-calls