On 10/14/2011 10:11 AM, Pavel Roskin wrote:
On Sat, 08 Oct 2011 17:28:42 -0500 Larry Finger<Larry.Finger@xxxxxxxxxxxx> wrote:+static inline void htol16_buf(u16 *buf, unsigned int size) +{ + size /= 2; + while (size--) + *(__le16 *)(buf + size) = cpu_to_le16(*(buf + size)); }I'm not not sure compilers would optimize it out on little-endian systems. Perhaps you want a define that uses this code on big-endian systems and does nothing on little endian systems. Also, it would be nice to have a compile-time check that size is even. Or maybe size should be the number of 16-bit words, but then it would be better to call the argument "count" or something like that.
The patch was dropped. Even so, as this routine is found in brcmsmac, your comments warrant further discussion.
I am pretty sure that the compiler would optimize out the entire htol16_buf routine. After substitution for cpu_to_le16() on a little-endian system, the statement in the while loop becomes '*(buf + size) = *(buf + size)', which is certainly optimized away, as will the now empty while loop. The entire routine is reduced to 'size /= 2'. As this will have no effect on the external world, it will also be dropped leaving an empty htol16_buf(). I don't think any "#ifdef __BIG_ENDIAN ... #endif" statements are needed.
Your suggestion that the argument be renamed is good, but there is no need to check for an even number as the data in question come from 16-bit reads of the SPROM on the b43 device. That number of 16-bit quantities was multiplied by 2 to get the byte count before calling this routine. Of course, the routine should have been passed the number of 16-bit words, not the byte count. My second version would have done this.
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