GCC 4.0.1 i386 code generation issues

[Christer Palm]

Hello!
I have the following piece of C++ code which is supposed to make 16-bit 
integers from 8-bit lo-byte/hi-byte values in an super-efficient way. 
However I'm a bit puzzled about the (i386) code generated by GCC.

----------------------------------------------------------------------
#include <iostream>
#include <stdint.h>

class Int16
{
private:
         union {
                 uint_fast16_t i16;
                 uint8_t i8[sizeof (uint_fast16_t)];
         };
         static const int lo_byte_pos = 0;
         static const int hi_byte_pos = 1;

public:
         inline void set(uint_fast16_t v) { i16 = v; }
         inline void setL(uint8_t v) { i8[lo_byte_pos] = v; }
         inline void setH(uint8_t v) { i8[hi_byte_pos] = v; }
         inline uint_fast16_t get() { return i16; }
};

volatile uint8_t hi = 0x12, lo = 0x34;

int main()
{
         Int16 i;
         i.set(lo);
         i.setL(lo);
         i.setH(hi);

         std::cout << i.get() << std::endl;
}
----------------------------------------------------------------------

The i386 code generated by GCC 4.0.1 (-O3) for the interesting part of 
main() is:

         movb    lo, %dl
         movzbl  %dl, %edx
         movb    lo, %al
         movb    %al, %dl
         movb    hi, %al
         movzbl  %al, %eax
         movb    %al, %dh

Which is obviously a pretty good job, but what still disturbs me a bit is;

1/ That GCC fails to transform
	movb    lo, %al
         movb    %al, %dl
         movb    hi, %al
         movb    %al, %dh
into
	movb    lo, %dl
         movb    hi, %dh

2/ The completely unnecessary zero-extension of the argument to setH():
         movzbl  %al, %eax

The funny thing is that this does not happen for setL(), even though 
they have exactly the same declaration.


Unfortunately I haven't been able to try it with 4.1.0 yet...

Any way around this? Can anyone enlighten me on what's going on?

Cheers,
--
Christer Palm