static struct net_device *alloc_netdev(int sizeof_priv, const char *mask,
void (*setup)(struct net_device *))
{
struct net_device *dev;
int alloc_size;

/* ensure 32-byte alignment of the private area */
alloc_size = sizeof (*dev) + sizeof_priv + 31;

dev = (struct net_device *) kmalloc (alloc_size, GFP_KERNEL);
if (dev == NULL)
{
printk(KERN_ERR "alloc_dev: Unable to allocate device memory.\n");
return NULL;
}

memset(dev, 0, alloc_size);

if (sizeof_priv)
dev->priv = (void *) (((long)(dev + 1) + 31) & ~31);

setup(dev);
strcpy(dev->name, mask);

return dev;
}

the function of alloc_netdev is above.I'm disturbed about the sentence:
alloc_size = sizeof (*dev) + sizeof_priv + 31;
dev->priv = (void *) (((long)(dev + 1) + 31) & ~31);
what about this mean?and how can it ensure 32-byte alignment?thanks!

do out the bitwise math of this:

//round up x to next 4 byte boundary
x = 14;
x = (x + 3) & ~3

i could explain it, but the only way to 'get it' is to do out the bits.

I still not fully understand,can you explain it exactly?

01110 is 14
(01110 + 00011) is x + 3 = 10001 is 17
10001 & 11100 is 17 & ~3 = 10000 is 16

16 is the first multiple of 4 greater than 14. It's rounding 14 up to the next 4 byte boundary (16). But it works with any number obviously. Not just 14

alloc_size = sizeof (*dev) + sizeof_priv + 31;
like this alloc_size,who can tell me why this need to add 31?? and
dev->priv = (void *) (((long)(dev + 1) + 31) & ~31);
I know what this mean, and anybody who can tell me this algorithm can be applied anywhere?

you can only use that technique for powers of two.